# Gravitation and Orbits help!

1. Nov 28, 2012

### phyzz

1. The problem statement, all variables and given/known data
A spacecraft orbiting the Sun uses its jet engine for slowing down its orbital rotation and changing the direction of its velocity. At the moment when the velocity is directed away from the Sun and has a magnitude of v = 30 km/s, the jet engine is switched off. At the same moment, the mass of the spacecraft is m = 1500 kg and the distance to the Sun is R = 150 x 10^9 m.

1) Find the distance at which the velocity of the spacecraft is zero
2) Now suppose that in addition to the jet engine, the spacecraft is equipped with a solar sail, which can be opened at the moment of time when the jet engine is switched off. A solar sail is a very large mirror, that reflects photons emitted by the Sun. The reflected photons change their momenta and thus generate a force directed away from the Sun: F = C/r^2 (where C = 1.2 x 10^17 Nm^2 is a constant and r is a distance to the Sun in meters). Is the power of the solar sail sufficient for this aircraft to leave the Solar system?
G = 6.67 x 10^-11, the mass of the Sun is M = 2.0 x 10^30kg

3. The attempt at a solution

1) I used F = ma
the force of gravitation being GmM/r^2 = ma

made a the subject of the formula

integrated to find v and used the information given to find the constant.

Then I integrated that v to get x, again finding a value for the constant using the relevant information.

I set the velocity equation equal to zero to find a value for t (ie the moment in time when the velocity is zero) and subbed that t into my x equation

I think my procedure is very coherent, at least it seems so to me, others are using the Conservation of Energy approach and now I'm getting really confused

2) I know that Power = work done/time

and that work done = (Force)(distance)

so I found the work done by the gravitation pull of the Sun by multiplying (GmM/x^2) by (x) where x is the distance I found previously and divided that by time (the same time I found when v = 0 previously)

I did exactly the same for the work done by the photons except this time I multiplied C/x^2 by x (x, again same as before) and divided by time (same as before).

I then verified the following inequality, if the power done by the gravitational pull of the sun > power done by photons then it doesn't leave the Solar System

I subbed in values for G and the M's and found that it indeed stays in the Solar System.

2. Nov 28, 2012

### haruspex

Should produce the same answer, but I would have thought using C of E would be much simpler.
Power is just the rate of energy transfer at some instant. Even if the power from the sun falls below that required to keep pushing the spacecraft further away, it may already have enough speed. Indeed, with a higher initial speed it might not have needed the sail at all.
OTOH, it's not immediately obvious that you can just look at the energy this time. The energy gained by the sail over time depends on how long it spent at different distances. It seems like you need to develop a differential equation for the motion. But as it happens, you don't. Look at the formula for the gravitational force at a given distance, and the one for the force on the sail. Notice anything?

3. Nov 28, 2012

### Staff: Mentor

For part (1), conservation of energy is the way to go. I don't see how you could integrate the acceleration over time, since you have the acceleration in terms of distance (at least not easily -- show your work if you've done it!).

A very simply approach is to write the expression for the specific total mechanical energy of a body in orbit:

$\xi = \frac{v^2}{2} - \frac{\mu}{r}$

where $\mu$ is the gravitational parameter for the Sun, and is equal to $G\,M_{sun}$.

In words it's the sum of the specific KE and PE, and is a constant of the motion of any body in free-fall in a conservative field.

What happens to the KE term when the velocity goes to zero? Can you then find r?