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

Answer 1

Answer:

Work done to shift the spacecraft from Earth surface to low Earth radius is given as

$W = \frac{GM_e m}{2R_e}$

Explanation:

As we know that spacecraft is at surface of Earth initially

So we will have

$U_i = -\frac{GM_e m}{R_e}$

now when it is at low radius Earth Orbit then we have

$U_f = -\frac{GM_e m}{2(R_e + h)}$

now we know that work done to shift the spacecraft from Earth Surface to Low earth orbit is change in total energy

$W = -\frac{GM_e m}{2(R_e + h)} + \frac{GM_e m}{R_e}$

so we have

$W = \frac{GM_e m}{R_e} (-\frac{1}{2(1 + h/R_e)} + 1)$

$W = \frac{GM_e m}{R_e} (-\frac{1}{2}(1 - \frac{h}{R_e}) + 1)$

$W = \frac{GM_e m}{2R_e}(1 + \frac{h}{R_e})$

since we know h << Re

so work done is given as

$W = \frac{GM_e m}{2R_e}$

Answer 2

The work done to launch the spacecraft in low earth orbit is  $\frac{GM_emh}{R_e(R_e+h)}$

Work-energy theorem:

The gravitational force is a conservative force. So, the total energy of the system must be conserved.

According to the work-energy theorem:

work done = - change in potential energy of the system.

W = -ΔPE

Initially, the potential energy of the satellite on the surface is:

PE = $-\frac{GM_em}{R_e}$

where m is the mass of the satellite

Let the orbit be at a height of h from the surface, so the potential energy in the orbit is :

PE' = $-\frac{GM_em}{R_e+h}$

ΔPE = PE'-PE

ΔPE = $-GM_em(\frac{1}{R_e+h}-\frac{1}{R_e})$

$\Delta PE=-\frac{GM_emh}{R_e(R_e+h)}$

Now work done:

W = - ΔPE

Thus,

$W=\frac{GM_emh}{R_e(R_e+h)}$

Learn more about work energy theorem:

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