# Kinetic energy of space probe launched from planet

1. Apr 4, 2012

### 1MileCrash

1. The problem statement, all variables and given/known data

Zero, a hypothetical planet, has a mass of 3.0 x 10^23 kg, a radius of 3.0 x 10^6 m, and no atmosphere. A 17 kg space probe is to be launched vertically from its surface.

If the probe is launched with an initial kinetic energy of 5.0 x 10^7 J, what will be its kinetic energy when it is 4.0 x 10^6 m from the center of Zero?

2. Relevant equations

3. The attempt at a solution

I figure that mechanical energy is conserved, and so

K1 + U1 = K2 + U2

With U given by GmM/r.

So

$K_{2} = K_{1} + GmM(r^{-1}_{2} - r^{-1}_{1})$

With r1 = 3.0x10^6, and r2 = 3.0x10^6 + 4.0x10^6 = 7.0x10^6

Which gives me the negative value -14794285.71 for K2.

I have no idea what is wrong with the equation I'm using.

Any ideas?

2. Apr 4, 2012

### SHISHKABOB

gravitational potential energy is negative

also I think that your r2 is a bit too big

3. Apr 4, 2012

### 1MileCrash

Yes, so

K1 - (GmM/r1) = K2 - (GmM/r2)

And

K2 = K1 - (GmM/r1) + (GmM/r2)
K2 = K1 + (GmM/r2) - (GmM/r1)
K2 = K1 + GMm((1/r2)-(1/r1))

Which is what I used..

4. Apr 4, 2012

### SHISHKABOB

I did

K2 = K1 - U1 + U2
K2 = K1 -(U1 - U2)
K2 = K1 -GmM((1/r1) - (1/r2))

and got 21652500 J

I also got the same number when I used your method

I think the problem is your value for r2