Part A) Which takes more energy per kilogram of satellite: launching the satellite from the...

Question:

Part A) Which takes more energy per kilogram of satellite: launching the satellite from the ground to a height of 1600 km above the ground or placing the satellite into orbit once it has reached that altitude? Part B) Repeat previous part for an orbit 3200 km above the ground. Part C) Repeat previous for an orbit 4800 km above the ground.

Circular Orbit

For a stable circular orbit, the force due to gravity is equal to the centripetal force neccessary to maintain that path.

For a given radius and tangential velocity, the centripetal acceleration is:

{eq}a_c=\frac{v_t^2}{r} {/eq}

This corresoponds to a centripetal force of:

{eq}F_c=ma_c=m\frac{v_t^2}{r} {/eq}

Newton's Law of Universal Gravitation gives the gravitational force as:

{eq}F_g=G\frac{Mm}{r^2} {/eq}

In the case of orbit, M is the mass being orbitted by m, and r is the distance between the centers of mass. Set this equal to the centripetal force.

{eq}G\frac{Mm}{r^2}=m\frac{v_t^2}{r} {/eq}

We can now solve for the neccessary orbital velocity.

{eq}v_t=\sqrt{\frac{GM}{r}} {/eq}

To determine the energy needed to reach a certain altitude, we need to integrate the gravitational force acting on the mass across the distance from the surface of the Earth to the desired orbit (the radius of the Earth plus the orbital altitude). The is the work done by gravity on the mass, which is equal in magnitude to the energy needed.

{eq}W_g=\int_{R_E}^{RE+A}{G\frac{Mm}{r^2} \ dr}=-G\frac{Mm}{r}|_{R_E}^{RE+A}=GMm(\frac{1}{R_E+A}-\frac{1}{RE}) {/eq}

The launch energy per kilogram is then given by dividing out the launched mass and inverting the sign:

{eq}\frac{LE}{kg}=GM(\frac{1}{RE}-\frac{1}{RE+A}) {/eq}

Now to determine the energy needed to place it in orbit at that altitude. Assume the object has attained that altitude exactly and has zero velocity. Rotational kinetic energy is given by:

{eq}KE_R=\frac{1}{2}I\omega^2 {/eq}

For a point mass, the moment of inertia is:

{eq}I=mr^2 {/eq}

From tangential velocity, angular velocity is given by:

{eq}\omega=\frac{v_t}{r} {/eq}

Using the expression for orbital velocity, we have rotational kinetic energy as:

{eq}KE_R=\frac{1}{2}mr^2(\frac{\sqrt{\frac{GM}{r}}}{r})^2=\frac{1}{2}m\frac{GM}{r} {/eq}

The orbital energy per kilogram for m is then, note the radius will be the radius of earth plus the orbital altitude:

{eq}\frac{OE}{kg}=\frac{GM}{2}\frac{1}{R_E+A} {/eq}

So we need to determine the inequality for:

{eq}GM(\frac{1}{RE}-\frac{1}{RE+A}) \ ? \ \frac{GM}{2}\frac{1}{R_E+A} {/eq}

Let's clean this up a bit:

{eq}\frac{1}{RE}-\frac{1}{RE+A} \ ? \ \frac{1}{2(R_E+A)} {/eq}

Taking the Earth's radius to be 6371 km, now check for each of the three given altitudes. Note we are no longer looking at the energies, just the relative values of them.

• A)

{eq}\frac{1}{6371}-\frac{1}{6371+1600} \ ? \ \frac{1}{2(6371+1600)} {/eq}

{eq}.000032 < .000063 {/eq}

More energy needed to establish orbit.

• B)

{eq}\frac{1}{6371}-\frac{1}{6371+3200} \ ? \ \frac{1}{2(6371+3200)} {/eq}

{eq}.000052 = .000052 {/eq}

Energies are equal

• C)

{eq}\frac{1}{6371}-\frac{1}{6371+4800} \ ? \ \frac{1}{2(6371+4800)} {/eq}

{eq}.000067 > .000045 {/eq}

More energy to reach that altitude