How Do You Calculate the Total Energy of a Geosynchronous Satellite?

Click For Summary
SUMMARY

The total energy of a geosynchronous satellite with a mass of 1500 kg, orbiting Earth at a height of 325 km and an orbital speed of 5000 m/s, cannot be accurately calculated due to inconsistencies in the provided parameters. A geosynchronous satellite must orbit at approximately 40,000 km with a speed of about 3000 m/s. The correct formulas to use for calculating total energy include gravitational potential energy (U = -GMm/r) and kinetic energy (Ek = 1/2 mv^2), leading to a total energy (Et = Ek + U) of -1.875 x 10^10 J when adjusted for a hypothetical mass of Earth.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy formulas
  • Familiarity with Newton's law of universal gravitation
  • Knowledge of orbital mechanics and geosynchronous orbits
  • Basic proficiency in algebra and unit conversions
NEXT STEPS
  • Study the principles of orbital mechanics, focusing on geosynchronous orbits
  • Learn about gravitational potential energy calculations using U = -GMm/r
  • Explore kinetic energy calculations in the context of satellite motion
  • Investigate the implications of varying gravitational constants at different altitudes
USEFUL FOR

Aerospace engineers, astrophysicists, physics students, and anyone involved in satellite design or orbital dynamics will benefit from this discussion.

ND3G
Messages
79
Reaction score
0
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
 
Physics news on Phys.org
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)
 
Last edited:
Thanks Chi, much appreciated
 
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.
 
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.
 
Last edited:
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:
ND3G said:
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 don't 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 don't 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?
 
cdaisy said:
I used 2 different formulas to find Ek, and got a different number from yours...

I don't 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 don't 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
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
12K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 21 ·
Replies
21
Views
3K