Ttdfjt

Does a vocal melody have any rhythmic responsibility to the underlying arrangement in pop music?

Prime sieve in Python

I found a password with hashcat, but it doesn't work

Can I enter the UK for 24 hours from a Schengen area, holding an Indian passport?

What determines the direction in which motor proteins go?

How do I farm creepers for XP without them exploding?

How to remove this component from PCB

Am I legally required to provide a (GPL licensed) source code even after a project is abandoned?

What can I do with a research project that is my university’s intellectual property?

Can White Castle?

Constitutionality of U.S. Democratic Presidential Candidate's Supreme Court Suggestion

What is the origin of Scooby-Doo's name?

How large would a mega structure have to be to host 1 billion people indefinitely?

Should an enameled cast iron pan be seasoned?

Understanding the reasoning of the woman who agreed with Shlomo to "cut the baby in half"

How does DC work with natural 20?

Will generated tokens be progressively stronger when using Cathar's Crusade and Sorin, Grim Nemesis?

`-` in tar xzf -

Designing a magic-compatible polearm

What is the highest voltage from the power supply a Raspberry Pi 3 B can handle without getting damaged?

How do I professionally let my manager know I'll quit over an issue?

Shooting someone's past self using special relativity

What is the meaning of "понаехать"?

If the Dragon's Breath spell is cast on a familiar, does it use the wizard's DC or familiar's DC?

Can artificial satellite positions affect tides?

What are the benefits of tides? Should we look for them on exoplanets?Why is momentum transferred to the moon?Would an asteroid collision affect Moon's orbit, and what consequence would that have for Earth?Tides on other bodiesCould daily variations of weight on Earth really be 0.003%?Could the Galilean moons tidally lock Jupiter?The sorting of perturbational effects by the power

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

14

3

$begingroup$

The Moon's position affects tides. So, is there any possibility for man-made satellite positions to also affect tides?

tides

share|improve this question

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Related: Do flights affect tides?
  $endgroup$
  – WBT
  Jun 12 at 23:36
  

  
  
  
  

add a comment | 

14

3

$begingroup$

The Moon's position affects tides. So, is there any possibility for man-made satellite positions to also affect tides?

tides

share|improve this question

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Related: Do flights affect tides?
  $endgroup$
  – WBT
  Jun 12 at 23:36
  

  
  
  
  

add a comment | 

$begingroup$

The Moon's position affects tides. So, is there any possibility for man-made satellite positions to also affect tides?

tides

share|improve this question

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

$endgroup$

share|improve this question

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

share|improve this question

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

edited Jun 12 at 19:07

RonJohn

249111

edited Jun 12 at 19:07

RonJohn

249111

asked Jun 12 at 4:56

namename

12419

asked Jun 12 at 4:56

namename

12419

3 Answers
3

active

oldest

votes

48

$begingroup$

Note: this is a 'Fermi estimate' answer, so I'm going to round off and ignore minor effects.

Tides are caused by gravity, and gravity is a really weak force. The mass of the Moon is 7 x 10²² kg, and it causes tides where the difference between water levels is on the order of 10 m.

Satellites are on the order of 10³ kg, and they're 10³ times closer, so the tidal effect caused by the average satellite is 10¹³ times smaller, and that's too small to be measurable.

Another factor is the number of satellites: 1000 sats in orbits evenly distributed around Earth will cancel out each other's tidal effects.

share|improve this answer

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

$endgroup$

• 
  
  
  

  
  30
  

  
  
  
  
  
  

  
  $begingroup$
  Another small correction: the amplitude of tides is proportional not to the strength of the gravity force, which is proportional to $r^-2$, but to its gradient, which is proportional to $r^-3$. So a satellite's tides would be "only" about $10^10$ times smaller than the Moon's.
  $endgroup$
  – Litho
  Jun 12 at 7:50
  
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer, but tides are actually only about 0.5m. Everything bigger than that is caused by the water sloshing around in the ocean bed – or, if you want to put it more respectably, by the rhythm of the tidal pull exciting a resonance in the given body of water.
  $endgroup$
  – Martin Kochanski
  Jun 12 at 10:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It might help to be clearer that a fact of $10^3$ in distance means a factor of $10^9$ in tidal effect.
  $endgroup$
  – dmckee
  Jun 12 at 19:16
  

  
  
  
  


• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  $begingroup$
  I just did the numbers quickly, but I think Pluto has a substantially larger effect on the tides than a satelite does.
  $endgroup$
  – Cort Ammon
  Jun 12 at 20:33
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Airplanes are both much heavier than satellites (excepting ISS) and a lot closer. So Boeing/Airbus tides should be 100 - 1000 times higher than even the ISS ones.
  $endgroup$
  – IMil
  Jun 12 at 23:31
  

  
  
  
  

 | 
show 7 more comments

6

$begingroup$

Everything that Hobbes says in his answer is correct, but doesn't seem to me to give a clear answer, which is YES they do affect tides, though not to a measurable extent.

When they were launched, that had a tiny effect on the earths rotation, though not particularly relevant here.

Every single one has exactly the same effect as the moon does, but to a far smaller extent, due to the tiny mass. The totality of artificial satellites, including the ISS, are NOT uniformly distributed, so they do not entirely cancel each other out.

share|improve this answer

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It's like asking "Can a pea raise a truck?" That answer is yes, but the net effect is so small as to be considered 0.
  $endgroup$
  – Machavity
  Jun 13 at 14:44
  

  
  
  
  

add a comment | 

3

$begingroup$

Let's do one small calculation.

The mass of the Moon ~ 7x10^22 kg, roughly 4x10^8 m away.

Mass of a really large satellite (ISS) ~ 5x10^6 kg roughly 4x10^5 m away.

The magnitude of the tide will be proportional to the mass of the orbiting object, and to first order, proportional to the cube of the distance.

For the Moon, M/(r^3) is roughly 0.001

For the ISS, M/(r^3) is roughly 8x10^-11

So, the ISS raises a tide that is around Seven orders (~1x10^7) of magnitude weaker than that of the Moon.

share|improve this answer

edited Jun 14 at 18:43

answered Jun 13 at 17:15

namename

12419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proportional to the cube of the distance, as others have pointed out.
  $endgroup$
  – TonyK
  Jun 13 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, in this case, it is proportional to the inverse square of the distance.
  $endgroup$
  – name
  Jun 14 at 2:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Stating it twice doesn't make it true.
  $endgroup$
  – TonyK
  Jun 14 at 8:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ya, now I understood, will correct it. Thanks!
  $endgroup$
  – name
  Jun 14 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surely the magnitiude will be inversely proportional to distance??
  $endgroup$
  – Mike Brockington
  yesterday
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f36702%2fcan-artificial-satellite-positions-affect-tides%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

48

$begingroup$

Note: this is a 'Fermi estimate' answer, so I'm going to round off and ignore minor effects.

Tides are caused by gravity, and gravity is a really weak force. The mass of the Moon is 7 x 10²² kg, and it causes tides where the difference between water levels is on the order of 10 m.

Satellites are on the order of 10³ kg, and they're 10³ times closer, so the tidal effect caused by the average satellite is 10¹³ times smaller, and that's too small to be measurable.

Another factor is the number of satellites: 1000 sats in orbits evenly distributed around Earth will cancel out each other's tidal effects.

share|improve this answer

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

$endgroup$

• 
  
  
  

  
  30
  

  
  
  
  
  
  

  
  $begingroup$
  Another small correction: the amplitude of tides is proportional not to the strength of the gravity force, which is proportional to $r^-2$, but to its gradient, which is proportional to $r^-3$. So a satellite's tides would be "only" about $10^10$ times smaller than the Moon's.
  $endgroup$
  – Litho
  Jun 12 at 7:50
  
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer, but tides are actually only about 0.5m. Everything bigger than that is caused by the water sloshing around in the ocean bed – or, if you want to put it more respectably, by the rhythm of the tidal pull exciting a resonance in the given body of water.
  $endgroup$
  – Martin Kochanski
  Jun 12 at 10:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It might help to be clearer that a fact of $10^3$ in distance means a factor of $10^9$ in tidal effect.
  $endgroup$
  – dmckee
  Jun 12 at 19:16
  

  
  
  
  


• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  $begingroup$
  I just did the numbers quickly, but I think Pluto has a substantially larger effect on the tides than a satelite does.
  $endgroup$
  – Cort Ammon
  Jun 12 at 20:33
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Airplanes are both much heavier than satellites (excepting ISS) and a lot closer. So Boeing/Airbus tides should be 100 - 1000 times higher than even the ISS ones.
  $endgroup$
  – IMil
  Jun 12 at 23:31
  

  
  
  
  

 | 
show 7 more comments

48

$begingroup$

Note: this is a 'Fermi estimate' answer, so I'm going to round off and ignore minor effects.

Tides are caused by gravity, and gravity is a really weak force. The mass of the Moon is 7 x 10²² kg, and it causes tides where the difference between water levels is on the order of 10 m.

Satellites are on the order of 10³ kg, and they're 10³ times closer, so the tidal effect caused by the average satellite is 10¹³ times smaller, and that's too small to be measurable.

Another factor is the number of satellites: 1000 sats in orbits evenly distributed around Earth will cancel out each other's tidal effects.

share|improve this answer

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

$endgroup$

• 
  
  
  

  
  30
  

  
  
  
  
  
  

  
  $begingroup$
  Another small correction: the amplitude of tides is proportional not to the strength of the gravity force, which is proportional to $r^-2$, but to its gradient, which is proportional to $r^-3$. So a satellite's tides would be "only" about $10^10$ times smaller than the Moon's.
  $endgroup$
  – Litho
  Jun 12 at 7:50
  
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer, but tides are actually only about 0.5m. Everything bigger than that is caused by the water sloshing around in the ocean bed – or, if you want to put it more respectably, by the rhythm of the tidal pull exciting a resonance in the given body of water.
  $endgroup$
  – Martin Kochanski
  Jun 12 at 10:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It might help to be clearer that a fact of $10^3$ in distance means a factor of $10^9$ in tidal effect.
  $endgroup$
  – dmckee
  Jun 12 at 19:16
  

  
  
  
  


• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  $begingroup$
  I just did the numbers quickly, but I think Pluto has a substantially larger effect on the tides than a satelite does.
  $endgroup$
  – Cort Ammon
  Jun 12 at 20:33
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Airplanes are both much heavier than satellites (excepting ISS) and a lot closer. So Boeing/Airbus tides should be 100 - 1000 times higher than even the ISS ones.
  $endgroup$
  – IMil
  Jun 12 at 23:31
  

  
  
  
  

 | 
show 7 more comments

$begingroup$

Note: this is a 'Fermi estimate' answer, so I'm going to round off and ignore minor effects.

Tides are caused by gravity, and gravity is a really weak force. The mass of the Moon is 7 x 10²² kg, and it causes tides where the difference between water levels is on the order of 10 m.

Satellites are on the order of 10³ kg, and they're 10³ times closer, so the tidal effect caused by the average satellite is 10¹³ times smaller, and that's too small to be measurable.

Another factor is the number of satellites: 1000 sats in orbits evenly distributed around Earth will cancel out each other's tidal effects.

share|improve this answer

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

$endgroup$

share|improve this answer

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

edited Jun 13 at 19:23

Jan Doggen

91811124

edited Jun 13 at 19:23

Jan Doggen

91811124

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

answered Jun 12 at 6:35

HobbesHobbes

99.2k2288442

6

$begingroup$

Everything that Hobbes says in his answer is correct, but doesn't seem to me to give a clear answer, which is YES they do affect tides, though not to a measurable extent.

When they were launched, that had a tiny effect on the earths rotation, though not particularly relevant here.

Every single one has exactly the same effect as the moon does, but to a far smaller extent, due to the tiny mass. The totality of artificial satellites, including the ISS, are NOT uniformly distributed, so they do not entirely cancel each other out.

share|improve this answer

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It's like asking "Can a pea raise a truck?" That answer is yes, but the net effect is so small as to be considered 0.
  $endgroup$
  – Machavity
  Jun 13 at 14:44
  

  
  
  
  

add a comment | 

6

$begingroup$

Everything that Hobbes says in his answer is correct, but doesn't seem to me to give a clear answer, which is YES they do affect tides, though not to a measurable extent.

When they were launched, that had a tiny effect on the earths rotation, though not particularly relevant here.

Every single one has exactly the same effect as the moon does, but to a far smaller extent, due to the tiny mass. The totality of artificial satellites, including the ISS, are NOT uniformly distributed, so they do not entirely cancel each other out.

share|improve this answer

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It's like asking "Can a pea raise a truck?" That answer is yes, but the net effect is so small as to be considered 0.
  $endgroup$
  – Machavity
  Jun 13 at 14:44
  

  
  
  
  

add a comment | 

$begingroup$

Everything that Hobbes says in his answer is correct, but doesn't seem to me to give a clear answer, which is YES they do affect tides, though not to a measurable extent.

When they were launched, that had a tiny effect on the earths rotation, though not particularly relevant here.

Every single one has exactly the same effect as the moon does, but to a far smaller extent, due to the tiny mass. The totality of artificial satellites, including the ISS, are NOT uniformly distributed, so they do not entirely cancel each other out.

share|improve this answer

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

$endgroup$

share|improve this answer

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

answered Jun 13 at 11:06

Mike BrockingtonMike Brockington

2373

3

$begingroup$

Let's do one small calculation.

The mass of the Moon ~ 7x10^22 kg, roughly 4x10^8 m away.

Mass of a really large satellite (ISS) ~ 5x10^6 kg roughly 4x10^5 m away.

The magnitude of the tide will be proportional to the mass of the orbiting object, and to first order, proportional to the cube of the distance.

For the Moon, M/(r^3) is roughly 0.001

For the ISS, M/(r^3) is roughly 8x10^-11

So, the ISS raises a tide that is around Seven orders (~1x10^7) of magnitude weaker than that of the Moon.

share|improve this answer

edited Jun 14 at 18:43

answered Jun 13 at 17:15

namename

12419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proportional to the cube of the distance, as others have pointed out.
  $endgroup$
  – TonyK
  Jun 13 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, in this case, it is proportional to the inverse square of the distance.
  $endgroup$
  – name
  Jun 14 at 2:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Stating it twice doesn't make it true.
  $endgroup$
  – TonyK
  Jun 14 at 8:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ya, now I understood, will correct it. Thanks!
  $endgroup$
  – name
  Jun 14 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surely the magnitiude will be inversely proportional to distance??
  $endgroup$
  – Mike Brockington
  yesterday
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Let's do one small calculation.

The mass of the Moon ~ 7x10^22 kg, roughly 4x10^8 m away.

Mass of a really large satellite (ISS) ~ 5x10^6 kg roughly 4x10^5 m away.

The magnitude of the tide will be proportional to the mass of the orbiting object, and to first order, proportional to the cube of the distance.

For the Moon, M/(r^3) is roughly 0.001

For the ISS, M/(r^3) is roughly 8x10^-11

So, the ISS raises a tide that is around Seven orders (~1x10^7) of magnitude weaker than that of the Moon.

share|improve this answer

edited Jun 14 at 18:43

answered Jun 13 at 17:15

namename

12419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proportional to the cube of the distance, as others have pointed out.
  $endgroup$
  – TonyK
  Jun 13 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, in this case, it is proportional to the inverse square of the distance.
  $endgroup$
  – name
  Jun 14 at 2:25
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Stating it twice doesn't make it true.
  $endgroup$
  – TonyK
  Jun 14 at 8:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ya, now I understood, will correct it. Thanks!
  $endgroup$
  – name
  Jun 14 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surely the magnitiude will be inversely proportional to distance??
  $endgroup$
  – Mike Brockington
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Let's do one small calculation.

The mass of the Moon ~ 7x10^22 kg, roughly 4x10^8 m away.

Mass of a really large satellite (ISS) ~ 5x10^6 kg roughly 4x10^5 m away.

The magnitude of the tide will be proportional to the mass of the orbiting object, and to first order, proportional to the cube of the distance.

For the Moon, M/(r^3) is roughly 0.001

For the ISS, M/(r^3) is roughly 8x10^-11

So, the ISS raises a tide that is around Seven orders (~1x10^7) of magnitude weaker than that of the Moon.

share|improve this answer

edited Jun 14 at 18:43

answered Jun 13 at 17:15

namename

12419

$endgroup$

share|improve this answer

edited Jun 14 at 18:43

answered Jun 13 at 17:15

namename

12419

answered Jun 13 at 17:15

namename

12419

answered Jun 13 at 17:15

namename

12419