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
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)"#

#= -(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.

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