What is the energy required to launch an m kg satellite from earth's surface in a circular orbit at an altitude of 2R? (R=radius of the earth)
2/3 mgR
- mgR
5/6 mgR
1/3 mgR
2/3 mgR- mgR
5/6 mgR1/3 mgR
1 Answer
Mar 10, 2018
3.
Explanation:
We need to use Law of conservation of Energy.
= -(GMm)/R +"KE"_r .......(1)
= -(GMm)/(3R) + 1/2 m (sqrt((GM)/(3R)))^2
= -(GMm)/(6R) ......(2)
Equating (1) and (2) and rearranging we get
"KE"_r = -(GMm)/(6R)+(GMm)/R
=>"KE"_r = (5GMm)/(6R)
Writing in terms of acceleration due to gravity
=>"KE"_r = 5/6mgR
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
Equating Centripetal force to gravitational force in the orbit of radius
(mv_0^2)/r=(GmM)/r^2
=>v_0= sqrt((GM)/r)
*Distances measured from centers of bodies, altitude from surface of earth.