Find speed of a satellite which orbits the moon near moon's surface?workings done,answer wrong nid pointers

Question

Find speed of a satellite which orbits the moon near moon's surface?What is the kinetic energy per unit mass of the satellite?
My workings are Fc(centripetal force)=Fg(gravitational force) so v=(GM/r)^(0.5)= 1680 m/s I got this correct but I took kinetic energy per unit mass of the satellite equals to gravitational field strength which is X=(GM/r) = 2.42x10^6 is my thinking correct? but the answer is 1.41x10^6J/kg.
Details given are radius of moon=1.74x10^6 m and mass of the moon=7.35x10^22 kg

Another extra unrelated question is that if the assumption is that gravitational potential at the surface of the earth to be zero, can we use the equation F=GMm/(r)^2?? Whats the meaning of the assumption? we use Energy=mgh instead??? Eg to find the gravitational field strength at p what do we do?????????

no wonder my teacher calls me a moron, i can't even solve simple things

Why am i so stupid? Lucky there is yahoo answers with kind people to help me

o by the way for the second part p means a a mass of 2kg at point p which is 3m above the surface of the earth

Answer 1

I would do this in terms of accelrrations rather than specific energy. The acceleration of gravity at the Moon's surface is 1.63 m/s^2. To achieve orbit, the centripital acceleration must match this. Centriptlal acceleration is v^2/r where v is the velocity and r is the radius of the path. The velocity then is v = squareroot(g x r) = 1690 m/sec. The specific kinetic energy is v^2/2 = 2 x g x r = 2870 kJ/kg.

I'm not sure what you mean by using the gravitaional field strength, but a correct approach along those lines is to integrate the gravitaitonal force from an infinite distance to the point in question. This will also yield the kinetic energy. Integrating (GM/r^2) gives (-2GM/r) applying the limits, at r= infinity it evaluates to zero so the total specific gravitational energy is (-2GM/r). Since energy is conserved, the sum of the kinetic and gravitational must not change. At infinite distance to gravitational energy is arbitrarily set at zero and the body is at rest so the total energy is zero. The sum of gravitational and kinetic must be zero in orbit as well so the kinetic is (2GM/r) = 2870 kJ/kg, exactly what resulted from the simple velocity approach.

Your second question is about gravity potential being zero at the Earth's surface. Normally the zero potential is assumed to be at infinite distance. As I showed above, this makes the calculations simple. You are free to choose any arbitrary radius but it will certainly complicate things.

There is a difference between F=GMm/r^2 and mgh. The mgh assumes that gravity is constant over the span of the problem. To do the true solution for energy, you need to do the integral I did above. and instead of h, you use limits on the integral of r0 and r0+h. For normal terrestrial applications, mgh is a lot simpler and gives an accurate answer. If you are trying to launch a rocket, mgh is nowhere near good enough.

Answer 2

You don't use the gravitational field to find kinetic energy per mass, but rather the velocity. KE = 0.5*mv^2. So KE/m = 0.5*v^2. It's just half the square of velocity.

I'm not sure what that assumption means, either. I think it's telling you to assume that the Earth's graviational effect on an object at the Moon's surface is negligible.

Answer 3

I would imagine a synchronised orbit around the Moon with the satellite placed on the far side of the Moon would do it. The next question of course would by why? The only reason I can think of for that would be a permanent imaging satellite of the lunar far side.

Answer 4

i didn't answer right away because i wanted to do my calculations slowly and accurately...the answer is quite simple: 992.772fps. 12.553fppsi.

i am not a scientist, however, i did stay at a holiday inn express last night!