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. 2/323 mgR
  2. mgR
  3. 5/656 mgR
  4. 1/313 mgR

1 Answer
A08
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)"Energy at earth's surface=PES+KEr(required for launch)

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

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

= -(GMm)/(3R) + 1/2 m (sqrt((GM)/(3R)))^2=−GMm3R+12m(√GM3R)2
= -(GMm)/(6R)=−GMm6R ......(2)

Equating (1) and (2) and rearranging we get

"KE"_r = -(GMm)/(6R)+(GMm)/RKEr=−GMm6R+GMmR
=>"KE"_r = (5GMm)/(6R)⇒KEr=5GMm6R

Writing in terms of acceleration due to gravity g=(GM)/R^2g=GMR2, we get

=>"KE"_r = 5/6mgR⇒KEr=56mgR

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

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

(mv_0^2)/r=(GmM)/r^2mv20r=GmMr2
=>v_0= sqrt((GM)/r)⇒v0=√GMr

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

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