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)

  1. 2/3 mgR
  2. mgR
  3. 5/6 mgR
  4. 1/3 mgR

1 Answer
Mar 10, 2018

3.

Explanation:

We need to use Law of conservation of Energy.

"Energy at earth's surface" = "PE"_S+ "KE"_r "(required for launch)"

= -(GMm)/R +"KE"_r .......(1)

"Energy in the orbit at altitude* of 2R" = "PE"_(o) +"KE"_o^$

= -(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 g=(GM)/R^2, we get

=>"KE"_r = 5/6mgR

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

"^$Calculations for orbital velocity.
Equating Centripetal force to gravitational force in the orbit of radius r

(mv_0^2)/r=(GmM)/r^2
=>v_0= sqrt((GM)/r)

*Distances measured from centers of bodies, altitude from surface of earth.