# Gravitational potential energy, orbital speed, binding energy.

1. Nov 25, 2012

### Lolagoeslala

1. The problem statement, all variables and given/known data
a satellite having a mass of 1800 kg orbits the earth at a distance of 6.3 x 10^5 m above the surface find the gravitational potential energy of the satellite while in orbit, the orbital speed and the binding satellite.

3. The attempt at a solution

gravitational potential energy
Eg = -GMm/Ro
Eg = -(6.67x10^-11 Nm^2/kg^2)(5.98x10^24kg)(1800kg)/(6.37x10^6m)+(6.3x10^5m)
Eg = -1.03 x 10^11 J

the orbital speed

Ek = 1/2(GMm/Ro)
Ek = 0.5 x -1.03 x 10^11 J
Ek = 5.128277145 x 10^10 J

Ek = 1/2mv^2
V = 7548.57 m/s

binding satellite
Eb = 1/2(GMm/Ro)
Eb = 0.5 x -1.03 x 10^11 J
Eb = 5.128277145 x 10^10 J

2. Nov 25, 2012

### haruspex

Whilst that is true, I would prefer to use the equivalence of the gravitational force and the centripetal force required to maintain the orbit. That seems to me to be a more fundamental principle.
I'm not sure about that. Does binding energy take into account the KE? Maybe it does.

3. Nov 25, 2012

### Lolagoeslala

You mean like this

m(v^2/Ro) = GMm/Ro^2 ????

Well binding energy is the energy required for the orbiting satellite to escape. So the total energy should be zero....

Well yes Ek is included...

Eg + Ek = Et1
- GMm/Ro + 1/2GMm/Ro = - 1/2 GMm/Ro

Et1 + Eb = Et2
- 1/2 GMm/Ro + Eb = 0 J
Eb = 1/2 GMm/Ro

4. Nov 25, 2012

Yes.
OK.