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Total Energy of a satellite

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Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot) with a mass of 1500kg, orbiting Earth at a height of 325km with an orbital speed of 5000m/s

This question is starting to drive me a little mad. First of all, the satellite can't be in a geosynchronous orbit AND traveling at 5000m/s 325km above the earth. The two just don't jive.

Secondly, I am completely at a loss as to what formulas I should be using.

Should I simply be using 1/2mv^2 + mgh (where g = equals the reduced gravity of 8.8756069 at that height) for the question or is it much more involved than that?

I've read over articles involving critical velocities, gravitational attraction, circular motion, and centrifugal force until my head is spinning.

Can some please please give me a nudge in the right direction?

Thanks
 

Answers and Replies

Chi Meson
Science Advisor
Homework Helper
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Oh my god.

Whoever framed this question won't have a clue as what the answer ought to be. You are totally correct in that a geosychrynous satellite won't be at that altitude or speed. Not only that, a satellite at the altitude given, won't have the speed that's given either.

And you are also correct in that the "mgh" formula is too simple. Since g will not be constant. I bet, however, this is what the sorry excuse for a teacher wants. I'd say give him/her the 1/2mv^2 + mgh answer, then explain how wrong the question is.

PS:

A geosynchronous satellite must be at an orbital radius of about 40,000,000 meters, and it's speed is about 3000 m/s. This is by using the period (T)of one earth day (in seconds), and satisfying three formulas:
F= GMm/r^2
F=(mv^2)/r
and v=(2(pi)r)/T

for any given T, there is only one v and one r that works.

The Total gravitational potential energy of an object is actually zero at a large distance from the earth, and is considered negative the closer you get to earth, using the equation U=-GMm/r (not r^2)
 
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Thanks Chi, much appreciated
 
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If you're really ambitious, you could assume that he's talking about some other planet coincidentally named "earth," and satisfy the three requirements that Chi mentioned.
 
81
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Yeah, I had considered that, I would rather work with the numbers they gave me seeing has it clearly is not based on the Earth I know.

Update: In order to make the equation work, I have changed the mass of my bizarro Earth to: 2.51311844078*10^24 kg and a full day only lasts 2.34 hours.
 
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Ok, after making the changes to the mass of earth as stated above I have:

Given:
G = 6.67*10^-11 (N*m^2)/kg^2
M = 2.51311844078*10^24 kg ***revised mass to fit the question***
m = 1500 kg
v = 5000 m/s

Required:
Et

Analysis:
F = (GMm)/r^2

Solution:
F = 5592.84116331 N

Analysis:
U = -(GMm)/r ***not r^2***

Solution:
U = -3.75*10^2 J

Analysis:
Ek = (1/2)(GMm/r)

Solution:
Ek = 1.875*10^10 J

Analysis:
Et = Ek +U

Solution:
Et = -1.875*10^10 J

Paraphrase:
The Total energy of the satellite is -1.875*10^10 J.

Does that look right to you guys? Thanks in advance
 
Last edited:
4
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Ok, after making the changes to the mass of earth as stated above I have:

Given:
G = 6.67*10^-11 (N*m^2)/kg^2
M = 2.51311844078*10^24 kg ***revised mass to fit the question***
m = 1500 kg
v = 5000 m/s

Required:
Et

Analysis:
F = (GMm)/r^2

Solution:
F = 5592.84116331 N

Analysis:
U = -(GMm)/r ***not r^2***

Solution:
U = -3.75*10^2 J

Analysis:
Ek = (1/2)(GMm/r)

Solution:
Ek = 1.875*10^10 J

Analysis:
Et = Ek +U

Solution:
Et = -1.875*10^10 J

Paraphrase:
The Total energy of the satellite is -1.875*10^10 J.

Does that look right to you guys? Thanks in advance
I used 2 different formulas to find Ek, and got a different number from yours...

I dont know what number you used for the radius, but I used 6378.1 km, which I found off google...

Ek = mv2
Ek = (1500kg)(5000m/s)2
Ek = 3.75 x 1010 J

Ek = -GMm/r
Ek = -[(6.67x10-11Nm2/kg2)(2.513x1024kg)(1500kg)]/6 378 100m
Ek = 2.514x1017Nm/6 379 100
Ek = 3.94x1010Nm

The answers are slightly different, possibly due to rounding... I dont understand why your equation is halved...

When I use the earth's real mass, I get Ek = 9.37 Nm, which does not match my first equation at all...

Also, is there any use for finding F?
 
4
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I used 2 different formulas to find Ek, and got a different number from yours...

I dont know what number you used for the radius, but I used 6378.1 km, which I found off google...

Ek = mv2
Ek = (1500kg)(5000m/s)2
Ek = 3.75 x 1010 J

Ek = -GMm/r
Ek = -[(6.67x10-11Nm2/kg2)(2.513x1024kg)(1500kg)]/6 378 100m
Ek = 2.514x1017Nm/6 379 100
Ek = 3.94x1010Nm

The answers are slightly different, possibly due to rounding... I dont understand why your equation is halved...

When I use the earth's real mass, I get Ek = 9.37 Nm, which does not match my first equation at all...

Also, is there any use for finding F?


oopss, nevermind this response, it makes sense
 

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