Find the satellite's altitude above the earth's surface

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Homework Help Overview

The discussion revolves around determining the altitude of a communication satellite above the Earth's surface, given its gravitational force and potential energy. The problem involves gravitational concepts and calculations related to orbital mechanics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the gravitational potential energy and gravitational force equations, questioning whether they can be equated. There is uncertainty about the correct interpretation of the radius used in calculations, particularly whether it should include the Earth's radius. Some participants attempt to relate force and potential energy through various equations.

Discussion Status

There is an ongoing exploration of the relationships between gravitational force and potential energy. Some participants have attempted calculations but express confusion over the results. Clarifications regarding the definitions and assumptions in the problem are being sought, with no clear consensus reached yet.

Contextual Notes

Participants note the potential misunderstanding regarding the reference point for gravitational potential energy and the need to clarify whether the radius of the Earth is included in the calculations. There is also mention of the problem's parameters possibly being miscopied or misinterpreted.

whitetiger
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When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r
 
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whitetiger said:
When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r
Forgive me I have just started here. You might be on the right track but (and my physics is rusty) r would likely include the Earth's radius as r should be measured from the centre of gravity.

Why not F=mg U=mgh U=Fh h=U/F

Sorry what does this (below) mean?
""and the earth-satellite gravitational potential energy (relative to zero at infinite separation)""

i suspect this needs to be understood again as to whether the radius of Earth is included or not.
 
sirius0 said:
Forgive me I have just started here. You might be on the right track but (and my physics is rusty) r would likely include the Earth's radius as r should be measured from the centre of gravity.

Why not F=mg U=mgh U=Fh h=U/F

Sorry what does this (below) mean?
""and the earth-satellite gravitational potential energy (relative to zero at infinite separation)""

i suspect this needs to be understood again as to whether the radius of Earth is included or not.

Yes, I have tried F=mg U=mgh U=Fh h=U/F and substituting in 1.49×10^11 J/ 15.1 kN. But I've got the wrong answer.

Can someone comment more on this

Thank
 
I think we have to use the radius of the Earth to calculate for the satellite's altitude

So the above equation F = mgU = mgh U=Fh h= U/F is not correct
 
whitetiger said:
When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r
No, you can't "set the two equations equal to each other"! They measure different things and are not equal.

Have you copied the problem correctly? You have both gravitational force and potential energy equal to -1.49 x 10^11 (one in Newtons and the other in Joules). It is possible that they happen to be the same but rather peculiar!

Assuming that is correct, you can use the fact that Fg= (GMeM)/r^2= 1.49 x 10^11 to determine that GMeM= 1.49 x 10^11 r^2. Now put that value into GMeM/r= 1.49 x 10^11 and solve for r. Hmm, you seem to have, then,
1.49 x10^11r^2/r= 1.49 x 10^11 from which we determine that r= 1 m. That satellite is "orbiting" one meter more than 10^6 m deep in the earth! Check the problem again!
 

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