I assume you have calculus. g is changing, so we have to use calculus.
Work= INT Force*dx
= INT GM3m/distance^2 dx integrated from radEarth to 2500km+radEarth.
A 1300-kg space vehicle falls vertically from a height of 2500 km above the earth's
surface. Determine how much work is done by the force of gravity in bringing the vehicle
to the earth's surface.
2 answers
For such a large change in altitude, the change of the acceleratin of gravity as it falls must be taken into account.
The Earth's radius is 6370 km. The change in potential energy falling from altitude 2500 km to the surface is
G M m (1/6370*10^3 - 1/8870*10^3 m)
where m = 1300 kg, M is the mass of the Earth and G is the universal constant of gravity. If you use the fact that GM/Re^2 = g, where Re is the Earth's radius and g is the acceleration of gravity there, then the P.E. change can be written
delta PE = m g Re^2[1/Re - 1/8870*10^3]
= m g Re[1 - Re/(Re+2500*10^3)]
= 1300*9.8*6370*10^3[1 - (6370/8870)]
= 2.29*10^10 Joules
The Earth's radius is 6370 km. The change in potential energy falling from altitude 2500 km to the surface is
G M m (1/6370*10^3 - 1/8870*10^3 m)
where m = 1300 kg, M is the mass of the Earth and G is the universal constant of gravity. If you use the fact that GM/Re^2 = g, where Re is the Earth's radius and g is the acceleration of gravity there, then the P.E. change can be written
delta PE = m g Re^2[1/Re - 1/8870*10^3]
= m g Re[1 - Re/(Re+2500*10^3)]
= 1300*9.8*6370*10^3[1 - (6370/8870)]
= 2.29*10^10 Joules
3 answers