Calculate how much work is required to launch a spacecraft of mass mm from the surface of the earth (mass mEmE, radius RERE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RERE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 kmkm, much less than RERE

Work done = (1/2)[(Gmm_e)/(R_e)]

Explanation:

I've attached the explanations below.

Question:

(a) 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=6370km

.)

Ignore the kinetic energy that the spacecraft has on the ground due to the earth’s rotation. (b) Calculate the minimum amount of additional work required to move the space craft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement “In terms of energy, low earth orbit is halfway to the edge of the universe.”

(a)

(b)

(c) The work required to put the spacecraft in low orbit is the same as the work to place the spacecraft at a very great distance away from the Earth

Explanation:

Here we have

The energy required is given by

Total energy, E = K.E + Gained P.E.

Where:

K.E. = Kinetic Energy

P.E. = Potential, gravitational Energy

P.E. can be found by

Where:

m = Mass of spacecraft

= Mass of the Earth

= Radius of the Earth

h = Height of the space craft above the Earth

G = Universal gravitational constant

Therefore, at height ( + h), we have, P.E. = m×g× ( + h)

But, m×g = Force =

Therefore, P. E. =   ×( + h) =

Since P. E. tends to act in opposite direction to K.E. which is moving to a higher altitude, we have;

Total energy, E  given by

However, we note that the spacecraft is in orbit, therefore

We note that to keep the spacecraft in orbit, we have

Therefore,

Which gives,

, that is

Total energy, E becomes

b) Given that the energy of the spacecraft on Earth is given by,

= since v = 0

The work required to move the spacecrraft, W, to the near orbit was found as

Where  h << we have

So that

,

The minimum energy required to move the spacecraft to  a very great distance from the earth is given again by;

K.E. + Gain in P.E.

Here, since w require the minimum energy, then our v→0 and our

→ ∞

Hence we have;

becomes

Therefore, from we have

Which gives

the answer for apex is the lowest

the right answer is 0.3 meters per second