HSC 2003 Q17b (2 Viewers)

[SkimDawg]

A Satellite of mass 150kg is launched from Earth's surface into a uniform circular orbit of radius 7.5*10^6m.
a) Calculate the magnitude of the gravitational potential energy Ep of the satellite.
I got 8*10^9N, which I think is right. But then..

b) From this uniform circular orbit, the satellite can escape Earth's gravitational field whe its kinetic energy is equal to the magnitude of the gravitational potential energy. Use this relationship to calculate the escape velocity of the satellite.

Im assuming it has something to do with F = mv^2/r, but to be honest I Have no idea.

Any help would be greatly appreciated, thanks

 

[samwell]

(a) for potential energy use e=-GM1M2/r
(b) since they tell u that escape velovity is where the KE=PE then u use
M1v^2/2 = -(-GM1M2/r) since KE is equal but opposite in direction a gain in KE is a loss in PE
Then manipulate the eqn to get escape velocity= sqr root of 2GM2/r now that u noe the radius and the mass ucan simply manipulate the eqn.

another easier way would b to use ua part a result. since PE=-KE u use M1v^2/2=-(-8*10^9)
then u should also get the same answer. Hope this is understandable.

 

[ratcher0071]

A Satellite of mass 150kg is launched from Earth's surface into a uniform circular orbit of radius 7.5*10^6m.
a) Calculate the magnitude of the gravitational potential energy Ep of the satellite.
I got 8*10^9N, which I think is right. But then..

b) From this uniform circular orbit, the satellite can escape Earth's gravitational field whe its kinetic energy is equal to the magnitude of the gravitational potential energy. Use this relationship to calculate the escape velocity of the satellite.

Im assuming it has something to do with F = mv^2/r, but to be honest I Have no idea.

Any help would be greatly appreciated, thanks

a) E_p = - G m₁m₂ / r
= - (6.67*10^-11) (5.98*10²⁴ x 150) / 7.5*10⁶
= - (6.67*10^-11) (1.196*10²⁰)
= - 8*10⁹ N

b) v = sqrt (2GM /r)
v = sqrt ((2 x 6.67*10^-11 x 5.98*10²⁴) / 6.38*10⁶)
v = sqrt 125 036 363.6
v=11 181.966 m/s
v=11.2 km/s

 

[SkimDawg]

thanks guys, very helpful.