Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth.?

Question

Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass me , radius Re) and place it in a circular low earth orbit, that is, an orbit whose altitude above the earth's surface is much less than Re . (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than Re=6380 km .) You can ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
Express your answer in terms of the variables m,me, Re and appropriate constants.

B)Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets.
Express your answer in terms of the variables m,me ,Re , and appropriate constants.

Answer 1

A) Since we're talking an orbit that is arbitrarily close to the earth's surface, we can ignore the increase in potential energy (due to raising the spacecraft's altitude); and concentrate just on its kinetic energy. The kinetic energy equals the work done.

Use these equations:

KE = mv²/2 (formula for kinetic energy)

a = v²/Re (because it's going in a circle)

Fnet = G(Me)(m)/Re² (because gravity is the only force acting on it)

Fnet = ma (Newton's 2nd Law)

All of those equations have variables in common, so you can combine the equations. So, combine the equations and then solve for KE.

B) Here they want to know the energy required to lift it to a "very great" (essentially infinite) height. The work equation is:

Work = Force × Distance

In this case, the Force you're working against is gravity:

Force = G(Me)(m)/R²

The problem is, we can't just plug that directly into the Work equation; because the value of "R" changes continually as we're moving farther and farther away, and that means the Force is continually changing too. That means we have to use calculus, to add up all the "bits" of work done as we move from R=Re to R=infinity:

Work = ∫(Force·dR)
Work = ∫(G(Me)(m)/R²·dR)

Integrate that between the limits R=Re and R=infinity.

Answer 2

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(A) Suppose mass of spacecraft=m

velocity of spacecraft in a low earth orbit= v

radius of orbit = r

centripetal force=mv^2/r

Suppose mass of earth= Me

Universal gravitational constant= G

Gravitational force on spacecraft=GmMe/r^2

As gravitational force provides the centrpetal force,

mv^2/r=GmMe/r^2

mv^2=GmMe/r

multiplying by (1/2) on both sides

(1/2) mv^2 =(1/2) GmMe/r

Final Kinetic energy of spacecraft in orbit =(1/2) mv^2 =(1/2) GmMe/r

Final Kinetic energy of spacecraft in orbit=(1/2) GmMe/r

assuming initial kinetic energy of spacecraft on the surface of earth to be zero

Change in kinetic energy=(1/2) GmMe/r

work required to launch spacecraft=change in kinetic energy

work required to launch spacecraft=(1/2) GmMe / r

As altitude of orbit above the surface of earth is much less than radius Re of earth,radius of orbit=r=Re

work required to launch spacecraft=(1/2) GmMe / Re
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(B) The minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth=W

As total energy at very great distance=E=zero,

W +kinetic energy +potential energy = E=zero

W +(1/2) GmMe/Re+ ( -GmMe/Re)=E=0

W - (1/2) GmMe/Re = 0

W = (1/2) GmMe/Re

The minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth=W =(1/2) GmMe / Re
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