Question

A spacecraft of mass 1.00×103kg orbits the sun (mass 1.99×1030kg) in a circular orbit of radius 1.50×1011m (equal to the average distance from the sun to the earth). You wish to move the spacecraft into a smaller circular orbit around the sun of radius 1.08×1011m (equal to the average distance from the sun to Venus). In doing this, what will be the change in the spacecraft's kinetic energy? Neglect the gravitational pulls of the planets on the spacecraft. Express your answer with the appropriate units.

Answer 1

Answer:

Change in Kinetic Energy is $- 3.44\times 10^{11}\ J$

Solution:

As per the question:

Mass of space craft, $m = 1.00\times 10^{3}\ kg$

Mass of the sun, M = $1.99\times 10^{30}\ kg$

Radius of the orbit, R = $1.50\times 10^{11}\ m$

Radius of smaller orbit around the sun, r = $1.08\times 10^{11}\ m$

Now,

We know that the Gravitational Potential Energy is given by:

$U = - \frac{Gm_{1}m_{2}}{r}$

where

G = Gravitational constant

r = distance of the space craft from the center of the sun

Now,

Initial Gravitational Potential Energy is:

$U_{in} = - \frac{GMm}{R}$

$U_{in} = - \frac{6.67\times 10^{- 11}\times 1.99\times 10^{30}\times 1.00\times 10^{3}}{(1.50\times 10^{11})}$

$U_{in} = - 8.849\times 10^{11}\ J$

Now,

Final Gravitational Potential Energy is:

$U_{f} = - \frac{GMm}{r}$

$U_{in} = - \frac{6.67\times 10^{- 11}\times 1.99\times 10^{30}\times 1.00\times 10^{3}}{(1.08\times 10^{11})}$

$U_{f} = - 1.229\times 10^{12}\ J$

Change in the Gravitational Potential Energy equals the change in the kinetic energy of the space craft.

$\Delta U = U_{f} - U_{i}$

$\Delta U = - 1.229\times 10^{12} - (- 8.849\times 10^{11}) = - 3.44\times 10^{11}\ J$

Thus

$\Delta KE = - 3.44\times 10^{11}\ J$