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

  • Thread starter Thread starter whitetiger
  • Start date Start date
  • Tags Tags
    Altitude Surface
AI Thread Summary
To find a communication satellite's altitude above Earth's surface, the gravitational potential energy and gravitational force equations are discussed. The gravitational potential energy is given as -1.49×10^11 J, while the force is 15.1 kN. Participants clarify that the radius used in calculations should include Earth's radius, and there is confusion about equating the two different measurements. One user suggests using the gravitational force to derive the radius, but another points out a potential error in the problem setup. The discussion emphasizes the need for careful interpretation of the equations and values provided.
whitetiger
Messages
22
Reaction score
0
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
 
Physics news on Phys.org
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!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top