Magnitude of vector quantitiesPotential energy sign conventionsSeparating the potential energy of a system of particles.Why do particles of equal mass (with one at rest) undergoing elastic collisions scatter at only right angles?Solving Lagrangian equations of motion for two point-bodies with gravitational interactionDeriving an equation for the mass of a pendulum (Follow up)?Work Done by Gravitational ForceWhy does gravity becomes negative when finding the negative derivative of potential? (I know $F=-dU/dx$) Why is $F$ itself negative?What does it mean if the dot product of two vectors is negative?Work done to change circular orbit and orbital speedSign in $Delta U= -intvecFcdot dvecl$

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Magnitude of vector quantities


Potential energy sign conventionsSeparating the potential energy of a system of particles.Why do particles of equal mass (with one at rest) undergoing elastic collisions scatter at only right angles?Solving Lagrangian equations of motion for two point-bodies with gravitational interactionDeriving an equation for the mass of a pendulum (Follow up)?Work Done by Gravitational ForceWhy does gravity becomes negative when finding the negative derivative of potential? (I know $F=-dU/dx$) Why is $F$ itself negative?What does it mean if the dot product of two vectors is negative?Work done to change circular orbit and orbital speedSign in $Delta U= -intvecFcdot dvecl$













2












$begingroup$


The gravitational force in the vector form is defined as $$ vecF=-fracGMmboldsymbol r^3vecr$$
Many textbooks define its magnitude as $$F_g=-fracGMmr^2$$



However, in the derivation of the gravitational potential energy $U_g=-fracGMmr$ $$W_textby gravity=int_c vecF_textgrav cdot mboxd vecr=int_r1^r2 F mboxdr cos(180^circ)=-int_r1^r2 F mboxdr$$
$$=-int_r1^r2 fracGMmr^2 mboxdr=+fracGMmrBiggr|^r_2_r_1=fracGMmr_2-fracGMmr_1$$



where $r_2>r_1$



since $Delta U_g=-W_by.grav$



$$Delta U_g=U_2-U_1=GMm Big(frac1r_1-frac1r_2Big)$$



setting $r_2=infty$, $U_2=0$



$$-U_1=GMmfrac1r_1$$



Therefore $$U=-fracGMmr$$



which implies that the correct way of writing the magnitude should be $F=fracGMmr^2$ (i.e. no minus sign) since the direction has already been taken into account(by cos(180)) during the dot product operation.



So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?



I'm having the same trouble when dealing with the force exerted by a spring in Hooke's Law, e.g. in the derivation of the EPE:



$$W_s=int vecF_s cdot mboxdvecx=int_x_1^x2kx mboxdx cos(180^circ)=-int_x_1^x2kx mboxdx$$
Using $Delta U_s=-W_s$ will yield $U_s=frac12kx^2$, which again implies that $F_s=kx$ (rather than $F_s=-kx$)



But in other derivations like the effective spring constant of a certain combination, $F_s=-kx$ is used instead.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    I would recommend you to read the definition of change in potential energy of a system.
    $endgroup$
    – Unique
    Jun 23 at 12:29










  • $begingroup$
    @Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
    $endgroup$
    – EXINT
    Jun 23 at 13:54
















2












$begingroup$


The gravitational force in the vector form is defined as $$ vecF=-fracGMmboldsymbol r^3vecr$$
Many textbooks define its magnitude as $$F_g=-fracGMmr^2$$



However, in the derivation of the gravitational potential energy $U_g=-fracGMmr$ $$W_textby gravity=int_c vecF_textgrav cdot mboxd vecr=int_r1^r2 F mboxdr cos(180^circ)=-int_r1^r2 F mboxdr$$
$$=-int_r1^r2 fracGMmr^2 mboxdr=+fracGMmrBiggr|^r_2_r_1=fracGMmr_2-fracGMmr_1$$



where $r_2>r_1$



since $Delta U_g=-W_by.grav$



$$Delta U_g=U_2-U_1=GMm Big(frac1r_1-frac1r_2Big)$$



setting $r_2=infty$, $U_2=0$



$$-U_1=GMmfrac1r_1$$



Therefore $$U=-fracGMmr$$



which implies that the correct way of writing the magnitude should be $F=fracGMmr^2$ (i.e. no minus sign) since the direction has already been taken into account(by cos(180)) during the dot product operation.



So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?



I'm having the same trouble when dealing with the force exerted by a spring in Hooke's Law, e.g. in the derivation of the EPE:



$$W_s=int vecF_s cdot mboxdvecx=int_x_1^x2kx mboxdx cos(180^circ)=-int_x_1^x2kx mboxdx$$
Using $Delta U_s=-W_s$ will yield $U_s=frac12kx^2$, which again implies that $F_s=kx$ (rather than $F_s=-kx$)



But in other derivations like the effective spring constant of a certain combination, $F_s=-kx$ is used instead.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    I would recommend you to read the definition of change in potential energy of a system.
    $endgroup$
    – Unique
    Jun 23 at 12:29










  • $begingroup$
    @Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
    $endgroup$
    – EXINT
    Jun 23 at 13:54














2












2








2





$begingroup$


The gravitational force in the vector form is defined as $$ vecF=-fracGMmboldsymbol r^3vecr$$
Many textbooks define its magnitude as $$F_g=-fracGMmr^2$$



However, in the derivation of the gravitational potential energy $U_g=-fracGMmr$ $$W_textby gravity=int_c vecF_textgrav cdot mboxd vecr=int_r1^r2 F mboxdr cos(180^circ)=-int_r1^r2 F mboxdr$$
$$=-int_r1^r2 fracGMmr^2 mboxdr=+fracGMmrBiggr|^r_2_r_1=fracGMmr_2-fracGMmr_1$$



where $r_2>r_1$



since $Delta U_g=-W_by.grav$



$$Delta U_g=U_2-U_1=GMm Big(frac1r_1-frac1r_2Big)$$



setting $r_2=infty$, $U_2=0$



$$-U_1=GMmfrac1r_1$$



Therefore $$U=-fracGMmr$$



which implies that the correct way of writing the magnitude should be $F=fracGMmr^2$ (i.e. no minus sign) since the direction has already been taken into account(by cos(180)) during the dot product operation.



So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?



I'm having the same trouble when dealing with the force exerted by a spring in Hooke's Law, e.g. in the derivation of the EPE:



$$W_s=int vecF_s cdot mboxdvecx=int_x_1^x2kx mboxdx cos(180^circ)=-int_x_1^x2kx mboxdx$$
Using $Delta U_s=-W_s$ will yield $U_s=frac12kx^2$, which again implies that $F_s=kx$ (rather than $F_s=-kx$)



But in other derivations like the effective spring constant of a certain combination, $F_s=-kx$ is used instead.










share|cite|improve this question











$endgroup$




The gravitational force in the vector form is defined as $$ vecF=-fracGMmboldsymbol r^3vecr$$
Many textbooks define its magnitude as $$F_g=-fracGMmr^2$$



However, in the derivation of the gravitational potential energy $U_g=-fracGMmr$ $$W_textby gravity=int_c vecF_textgrav cdot mboxd vecr=int_r1^r2 F mboxdr cos(180^circ)=-int_r1^r2 F mboxdr$$
$$=-int_r1^r2 fracGMmr^2 mboxdr=+fracGMmrBiggr|^r_2_r_1=fracGMmr_2-fracGMmr_1$$



where $r_2>r_1$



since $Delta U_g=-W_by.grav$



$$Delta U_g=U_2-U_1=GMm Big(frac1r_1-frac1r_2Big)$$



setting $r_2=infty$, $U_2=0$



$$-U_1=GMmfrac1r_1$$



Therefore $$U=-fracGMmr$$



which implies that the correct way of writing the magnitude should be $F=fracGMmr^2$ (i.e. no minus sign) since the direction has already been taken into account(by cos(180)) during the dot product operation.



So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?



I'm having the same trouble when dealing with the force exerted by a spring in Hooke's Law, e.g. in the derivation of the EPE:



$$W_s=int vecF_s cdot mboxdvecx=int_x_1^x2kx mboxdx cos(180^circ)=-int_x_1^x2kx mboxdx$$
Using $Delta U_s=-W_s$ will yield $U_s=frac12kx^2$, which again implies that $F_s=kx$ (rather than $F_s=-kx$)



But in other derivations like the effective spring constant of a certain combination, $F_s=-kx$ is used instead.







newtonian-mechanics newtonian-gravity work potential-energy vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 23 at 18:49









Qmechanic

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asked Jun 23 at 8:04









EXINTEXINT

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  • 1




    $begingroup$
    I would recommend you to read the definition of change in potential energy of a system.
    $endgroup$
    – Unique
    Jun 23 at 12:29










  • $begingroup$
    @Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
    $endgroup$
    – EXINT
    Jun 23 at 13:54













  • 1




    $begingroup$
    I would recommend you to read the definition of change in potential energy of a system.
    $endgroup$
    – Unique
    Jun 23 at 12:29










  • $begingroup$
    @Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
    $endgroup$
    – EXINT
    Jun 23 at 13:54








1




1




$begingroup$
I would recommend you to read the definition of change in potential energy of a system.
$endgroup$
– Unique
Jun 23 at 12:29




$begingroup$
I would recommend you to read the definition of change in potential energy of a system.
$endgroup$
– Unique
Jun 23 at 12:29












$begingroup$
@Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
$endgroup$
– EXINT
Jun 23 at 13:54





$begingroup$
@Unique Yes and isn't $Delta U_g=-W_by.grav$ consistent with this definition?
$endgroup$
– EXINT
Jun 23 at 13:54











3 Answers
3






active

oldest

votes


















3












$begingroup$

The formula
$$F_G = - G fracM mr^2$$
is correct. The minus sign represents the fact that the gravitational force is attractive.



There is a small mistake in your calculation of $W_G$, when you substitute $F_G$ with its formula. The formula for $F_G$ contains a minus sign that should have cancelled the minus before the integral:
$$- int_r_1^r_2 F mboxdr = int_r_1^r_2 G fracM mr^2 mboxdr$$
I also suppose that, since you used $cos(180°)$, you are considering the case where $r_2$ > $r_1$.



From this you derive that
$$W_G = GMm left( frac1r_1 - frac1r_2 right)$$
and this is indeed positive, because you need energy to separate two objects that are attracting each other.



For a conservative force, such as the gravitational force or the elastic force, work is also defined as $W = - Delta U$, where $Delta U$ is the change in potential energy. Notice the minus sign here. There is a reason for it. Work, as I said above, is the energy needed to go from one place to another. It's positive when you need to give energy and negative when you obtain energy (usually kinetic energy). The potential on the other hand has a different interpretation, it's one component of the energy possessed by an object.



Since we are usually interested in the difference in potential energy, and not its absolute value, we have a certain freedom to choose when the potential energy is 0. Usually for the gravitational case 0 is chosen as the potential energy at infinite distance, while for the elastic case it's at the rest position ($x = 0$).



So if
$$W_G = G fracM mr_1 - G fracM mr_2$$
and
$$W_G = - Delta U = U(r_1) - U(r_2)$$
we obtain that
$$U(r) = - G fracM mr$$
No need to change the sign of $F$. And the same principle applies to the elastic case.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
    $endgroup$
    – EXINT
    Jun 23 at 11:25











  • $begingroup$
    @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
    $endgroup$
    – GRB
    Jun 23 at 12:05










  • $begingroup$
    But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
    $endgroup$
    – EXINT
    Jun 23 at 13:57











  • $begingroup$
    @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
    $endgroup$
    – GRB
    Jun 24 at 8:19


















2












$begingroup$

I would not provide the whole derivation for the expression of potential energy of a system but I would like to give you the correct reasoning for the signs which have to be taken into consideration while deriving the expressions of potential energy.please consider the definition below which may help you:-



The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system.



The negative sign will effect some equations you posted.






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$


    So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?




    The answer is yes, you just remember magnitude without negative sign as negative just suggest direction and says it is attractive force. So in whole physics to understand easily never put negative in front of magnitudes as they just suggest direction only.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The formula
      $$F_G = - G fracM mr^2$$
      is correct. The minus sign represents the fact that the gravitational force is attractive.



      There is a small mistake in your calculation of $W_G$, when you substitute $F_G$ with its formula. The formula for $F_G$ contains a minus sign that should have cancelled the minus before the integral:
      $$- int_r_1^r_2 F mboxdr = int_r_1^r_2 G fracM mr^2 mboxdr$$
      I also suppose that, since you used $cos(180°)$, you are considering the case where $r_2$ > $r_1$.



      From this you derive that
      $$W_G = GMm left( frac1r_1 - frac1r_2 right)$$
      and this is indeed positive, because you need energy to separate two objects that are attracting each other.



      For a conservative force, such as the gravitational force or the elastic force, work is also defined as $W = - Delta U$, where $Delta U$ is the change in potential energy. Notice the minus sign here. There is a reason for it. Work, as I said above, is the energy needed to go from one place to another. It's positive when you need to give energy and negative when you obtain energy (usually kinetic energy). The potential on the other hand has a different interpretation, it's one component of the energy possessed by an object.



      Since we are usually interested in the difference in potential energy, and not its absolute value, we have a certain freedom to choose when the potential energy is 0. Usually for the gravitational case 0 is chosen as the potential energy at infinite distance, while for the elastic case it's at the rest position ($x = 0$).



      So if
      $$W_G = G fracM mr_1 - G fracM mr_2$$
      and
      $$W_G = - Delta U = U(r_1) - U(r_2)$$
      we obtain that
      $$U(r) = - G fracM mr$$
      No need to change the sign of $F$. And the same principle applies to the elastic case.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
        $endgroup$
        – EXINT
        Jun 23 at 11:25











      • $begingroup$
        @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
        $endgroup$
        – GRB
        Jun 23 at 12:05










      • $begingroup$
        But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
        $endgroup$
        – EXINT
        Jun 23 at 13:57











      • $begingroup$
        @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
        $endgroup$
        – GRB
        Jun 24 at 8:19















      3












      $begingroup$

      The formula
      $$F_G = - G fracM mr^2$$
      is correct. The minus sign represents the fact that the gravitational force is attractive.



      There is a small mistake in your calculation of $W_G$, when you substitute $F_G$ with its formula. The formula for $F_G$ contains a minus sign that should have cancelled the minus before the integral:
      $$- int_r_1^r_2 F mboxdr = int_r_1^r_2 G fracM mr^2 mboxdr$$
      I also suppose that, since you used $cos(180°)$, you are considering the case where $r_2$ > $r_1$.



      From this you derive that
      $$W_G = GMm left( frac1r_1 - frac1r_2 right)$$
      and this is indeed positive, because you need energy to separate two objects that are attracting each other.



      For a conservative force, such as the gravitational force or the elastic force, work is also defined as $W = - Delta U$, where $Delta U$ is the change in potential energy. Notice the minus sign here. There is a reason for it. Work, as I said above, is the energy needed to go from one place to another. It's positive when you need to give energy and negative when you obtain energy (usually kinetic energy). The potential on the other hand has a different interpretation, it's one component of the energy possessed by an object.



      Since we are usually interested in the difference in potential energy, and not its absolute value, we have a certain freedom to choose when the potential energy is 0. Usually for the gravitational case 0 is chosen as the potential energy at infinite distance, while for the elastic case it's at the rest position ($x = 0$).



      So if
      $$W_G = G fracM mr_1 - G fracM mr_2$$
      and
      $$W_G = - Delta U = U(r_1) - U(r_2)$$
      we obtain that
      $$U(r) = - G fracM mr$$
      No need to change the sign of $F$. And the same principle applies to the elastic case.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
        $endgroup$
        – EXINT
        Jun 23 at 11:25











      • $begingroup$
        @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
        $endgroup$
        – GRB
        Jun 23 at 12:05










      • $begingroup$
        But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
        $endgroup$
        – EXINT
        Jun 23 at 13:57











      • $begingroup$
        @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
        $endgroup$
        – GRB
        Jun 24 at 8:19













      3












      3








      3





      $begingroup$

      The formula
      $$F_G = - G fracM mr^2$$
      is correct. The minus sign represents the fact that the gravitational force is attractive.



      There is a small mistake in your calculation of $W_G$, when you substitute $F_G$ with its formula. The formula for $F_G$ contains a minus sign that should have cancelled the minus before the integral:
      $$- int_r_1^r_2 F mboxdr = int_r_1^r_2 G fracM mr^2 mboxdr$$
      I also suppose that, since you used $cos(180°)$, you are considering the case where $r_2$ > $r_1$.



      From this you derive that
      $$W_G = GMm left( frac1r_1 - frac1r_2 right)$$
      and this is indeed positive, because you need energy to separate two objects that are attracting each other.



      For a conservative force, such as the gravitational force or the elastic force, work is also defined as $W = - Delta U$, where $Delta U$ is the change in potential energy. Notice the minus sign here. There is a reason for it. Work, as I said above, is the energy needed to go from one place to another. It's positive when you need to give energy and negative when you obtain energy (usually kinetic energy). The potential on the other hand has a different interpretation, it's one component of the energy possessed by an object.



      Since we are usually interested in the difference in potential energy, and not its absolute value, we have a certain freedom to choose when the potential energy is 0. Usually for the gravitational case 0 is chosen as the potential energy at infinite distance, while for the elastic case it's at the rest position ($x = 0$).



      So if
      $$W_G = G fracM mr_1 - G fracM mr_2$$
      and
      $$W_G = - Delta U = U(r_1) - U(r_2)$$
      we obtain that
      $$U(r) = - G fracM mr$$
      No need to change the sign of $F$. And the same principle applies to the elastic case.






      share|cite|improve this answer









      $endgroup$



      The formula
      $$F_G = - G fracM mr^2$$
      is correct. The minus sign represents the fact that the gravitational force is attractive.



      There is a small mistake in your calculation of $W_G$, when you substitute $F_G$ with its formula. The formula for $F_G$ contains a minus sign that should have cancelled the minus before the integral:
      $$- int_r_1^r_2 F mboxdr = int_r_1^r_2 G fracM mr^2 mboxdr$$
      I also suppose that, since you used $cos(180°)$, you are considering the case where $r_2$ > $r_1$.



      From this you derive that
      $$W_G = GMm left( frac1r_1 - frac1r_2 right)$$
      and this is indeed positive, because you need energy to separate two objects that are attracting each other.



      For a conservative force, such as the gravitational force or the elastic force, work is also defined as $W = - Delta U$, where $Delta U$ is the change in potential energy. Notice the minus sign here. There is a reason for it. Work, as I said above, is the energy needed to go from one place to another. It's positive when you need to give energy and negative when you obtain energy (usually kinetic energy). The potential on the other hand has a different interpretation, it's one component of the energy possessed by an object.



      Since we are usually interested in the difference in potential energy, and not its absolute value, we have a certain freedom to choose when the potential energy is 0. Usually for the gravitational case 0 is chosen as the potential energy at infinite distance, while for the elastic case it's at the rest position ($x = 0$).



      So if
      $$W_G = G fracM mr_1 - G fracM mr_2$$
      and
      $$W_G = - Delta U = U(r_1) - U(r_2)$$
      we obtain that
      $$U(r) = - G fracM mr$$
      No need to change the sign of $F$. And the same principle applies to the elastic case.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jun 23 at 9:21









      GRBGRB

      1,0491 gold badge10 silver badges24 bronze badges




      1,0491 gold badge10 silver badges24 bronze badges











      • $begingroup$
        Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
        $endgroup$
        – EXINT
        Jun 23 at 11:25











      • $begingroup$
        @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
        $endgroup$
        – GRB
        Jun 23 at 12:05










      • $begingroup$
        But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
        $endgroup$
        – EXINT
        Jun 23 at 13:57











      • $begingroup$
        @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
        $endgroup$
        – GRB
        Jun 24 at 8:19
















      • $begingroup$
        Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
        $endgroup$
        – EXINT
        Jun 23 at 11:25











      • $begingroup$
        @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
        $endgroup$
        – GRB
        Jun 23 at 12:05










      • $begingroup$
        But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
        $endgroup$
        – EXINT
        Jun 23 at 13:57











      • $begingroup$
        @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
        $endgroup$
        – GRB
        Jun 24 at 8:19















      $begingroup$
      Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
      $endgroup$
      – EXINT
      Jun 23 at 11:25





      $begingroup$
      Does $W_G$ denote the work done against gravity here? In my derivation $W_G$ is actually the work done by gravity (apologies for not clarifying....) which I thought is negative whenever $r_2$>$r_1$ . If $W_G$ is the work done BY gravity then is there still an error?
      $endgroup$
      – EXINT
      Jun 23 at 11:25













      $begingroup$
      @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
      $endgroup$
      – GRB
      Jun 23 at 12:05




      $begingroup$
      @EXINT The only difference between work done against gravity and by gravity is the sign of the result, the starting formula is the same. You obtain the wrong sign because of the mistake I corrected in the second formula in my answer. Gravitational work should always be positive when $r_2 > r_1$.
      $endgroup$
      – GRB
      Jun 23 at 12:05












      $begingroup$
      But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
      $endgroup$
      – EXINT
      Jun 23 at 13:57





      $begingroup$
      But shouldn't it be negative if $ boldsymbolr $ is the outward-pointing radial vector since the work done by gravity Is against the direction of motion $(r_2>r_1)$?
      $endgroup$
      – EXINT
      Jun 23 at 13:57













      $begingroup$
      @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
      $endgroup$
      – GRB
      Jun 24 at 8:19




      $begingroup$
      @EXINT That's not how work is defined. Work is the amount of energy that has to be invested in order to reach a goal.
      $endgroup$
      – GRB
      Jun 24 at 8:19











      2












      $begingroup$

      I would not provide the whole derivation for the expression of potential energy of a system but I would like to give you the correct reasoning for the signs which have to be taken into consideration while deriving the expressions of potential energy.please consider the definition below which may help you:-



      The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system.



      The negative sign will effect some equations you posted.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        I would not provide the whole derivation for the expression of potential energy of a system but I would like to give you the correct reasoning for the signs which have to be taken into consideration while deriving the expressions of potential energy.please consider the definition below which may help you:-



        The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system.



        The negative sign will effect some equations you posted.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          I would not provide the whole derivation for the expression of potential energy of a system but I would like to give you the correct reasoning for the signs which have to be taken into consideration while deriving the expressions of potential energy.please consider the definition below which may help you:-



          The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system.



          The negative sign will effect some equations you posted.






          share|cite|improve this answer









          $endgroup$



          I would not provide the whole derivation for the expression of potential energy of a system but I would like to give you the correct reasoning for the signs which have to be taken into consideration while deriving the expressions of potential energy.please consider the definition below which may help you:-



          The change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system.



          The negative sign will effect some equations you posted.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 23 at 12:27









          UniqueUnique

          1




          1





















              0












              $begingroup$


              So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?




              The answer is yes, you just remember magnitude without negative sign as negative just suggest direction and says it is attractive force. So in whole physics to understand easily never put negative in front of magnitudes as they just suggest direction only.






              share|cite|improve this answer











              $endgroup$

















                0












                $begingroup$


                So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?




                The answer is yes, you just remember magnitude without negative sign as negative just suggest direction and says it is attractive force. So in whole physics to understand easily never put negative in front of magnitudes as they just suggest direction only.






                share|cite|improve this answer











                $endgroup$















                  0












                  0








                  0





                  $begingroup$


                  So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?




                  The answer is yes, you just remember magnitude without negative sign as negative just suggest direction and says it is attractive force. So in whole physics to understand easily never put negative in front of magnitudes as they just suggest direction only.






                  share|cite|improve this answer











                  $endgroup$




                  So should I always use the expression of the magnitude without the negative sign because the direction of the force is only to be considered "later" in a sense that the negative sign should not be part of the magnitude?




                  The answer is yes, you just remember magnitude without negative sign as negative just suggest direction and says it is attractive force. So in whole physics to understand easily never put negative in front of magnitudes as they just suggest direction only.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jun 23 at 11:24









                  MarianD

                  9451 gold badge6 silver badges12 bronze badges




                  9451 gold badge6 silver badges12 bronze badges










                  answered Jun 23 at 8:17









                  sarthaksarthak

                  285 bronze badges




                  285 bronze badges



























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