1. Mar 19, 2006

### nick727kcin

this is the question:
A starship is circling a distant planet of radius R. The astronauts find that the acceleration due to gravity at their altitude is half the value at the planet's surface. How far above the surface are they orbiting? Your answer will be a multiple of R.

using a= (GM)/r^2, i found out that d=2R

this is wrong, though

2. Mar 19, 2006

### Da-Force

Explanation.

NOTES:
I will define 'a' as the planet's acceleration.
I will definie 'A' as the spaceship's acceleration.
I will define 'd' as their current altitude with respect to the planet.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

$$a_g = (G M) / r^2$$

The planet has a r=R, so the 'a' at the planet's surface is:

$$a_g = (G M) / R^2$$

They realize that their 'A' at their altitude is 'a/2'

$$A_g = (G M) / d^2$$
OR
$$(a/2)_g = (G M) / d^2$$

Let's look at this for a second. Everything in this equation is constant except for $$R^2$$ and $$d^2$$ and acceleration of gravity.

We can summarize this: $$a_g \sim 1/r^2$$

The acceleration of gravity is inversely proportional to r-squared. If we double the distance r, the acceleration is 1/4 than it was at a distance r.

So... solving for 'r'.

$$r^2 = 1/a_g$$
$$r = \sqrt{1/a_g}$$

So if the acceleration is 1/2

$$r = \sqrt{1/1/2}$$
$$r = \sqrt{2}$$

So there is your answer, mathematically... Now, having your answer as a multiple of r means ANY REAL number, not a whole number.

Last edited: Mar 19, 2006
3. Mar 19, 2006

### nick727kcin

thank you!

:!!)

4. Nov 22, 2006

### mawalker

i don't understand.... i understand how r = sqrt of 2, mathematically, but how does this equate to being a multiple of r?

5. Nov 22, 2006

### mawalker

nevermind...i see what was going on... tricky problem i entered the square root of 2 twice and now i understand that they ask (how far above)