How much work is required to lift a 700kg satellite to an altitude of 1x10^6m above the surface of the earth? The radius of the earth is 6.4x10^6m, its mass is 6x10^24 kg, and in these units the gravitational constant, G, is 6.67x10^-11

1 Answer
A08
Feb 16, 2018

There are two approaches through which this can be solved.

  1. #int\ vec(F_g) cdot vec(dr)# from initial to final location.
  2. By finding out difference in Gravitational Potential energy between two locations.

I opt for second approach.

Gravitational Potential Energy #U# possessed by an object of mass #m# in Earth's Gravitational Field is given by

#U= -(GMm)/r#
where #G# is Universal Gravitational constant, #M# is mass of earth and #r# is distance between the centers of earth and of the object.

Now Work done #W=# Change in Gravitational Potential Energy

#:.W = GPE("at the altitude " h) - GPE("surface")#
#=>W = -(G Mm)/[R+h] - (-(GMm)/R)#
where #R# is radius of earth
#=>W = G Mm[1/R-1/(R+h)]#

Inserting given values we get

#W = (6.67 xx 10^-11) (6 xx 10^24) (700) [1/(6.4 xx10^6) - 1/((6.4 xx 10^6) + (1 xx 10^6))]#
#=>W = 2.80xx 10^17 [1/(6.4 xx10^6) - 1/(7.4 xx 10^6)]#
#=>W = 2.80xx 10^11 [1/(6.4) - 1/(7.4 )]#
#=>W = 5.9 xx 10^9\ J#, rounded to one decimal place.

Related questions
Impact of this question
12971 views around the world
You can reuse this answer
Creative Commons License