# Gravitional Potential Energy Question

1. Nov 16, 2014

### squirrelschaser

1. The problem statement, all variables and given/known data
Planet 1 has mass 3M and radius R, while Planet 2 has mass 4Mand radius 2R. They are separated by center-to-center distance 8R. A rock is placed halfway between their centers at point O. It is released from rest. Ignore any motion of the planets.
The rock is released from rest at point O. Derive an expression for the speed v with which the rock crashes into a planet.

2. Relevant equations

KE = 1/2 *m *v^2
PE = GMm/R

3. The attempt at a solution

(3GMm/4R) -4GMm/4R = .5mV^2 + (3GMm/6R) - (4GMm/2R)

-1GMm/4R = (.5mV^2) - (3GMm/2R)

5GMm/4R = .5mV^2

5GMm/2R = mV^2

5GM/2R = V^2

2. Nov 16, 2014

### Simon Bridge

Note - how far is point O from each of the centers?

3. Nov 16, 2014

### TSny

Check the sign of the right hand side.

Why are some of your PE terms positive while others are negative?

4. Nov 16, 2014

### squirrelschaser

4R. IT's halfway between the center and center point. So 8R/2.

5. Nov 16, 2014

### squirrelschaser

Not sure. I arbitrarily chose the attraction to the left as positive (PE from the moon) and the attraction to the right as negative (PE from the Earth).

Should they all be negative?

6. Nov 16, 2014

### Simon Bridge

... hence the 4R in your equations?

You should not be arbitrary - use your understanding of physics. Note: potential energy is not a vector.

7. Nov 16, 2014

### squirrelschaser

?.

Well maybe not arbitrarily, I thought since the 2 PE are in opposite direction, one would be positive and one would be negative.

I will try this it with all the PE as negative values.

Last edited: Nov 16, 2014
8. Nov 16, 2014

### squirrelschaser

-(3GMm/4R) -4GMm/4R = .5mV^2 - (3GMm/6R) - (4GMm/2R)

-7GMm/4R = .5mV^2 - 15GMm/6R

9GMm/12R = .5V^2

9GM/12R = .5V^2

18GM/12R = V^2

sqrt(1.5GM/R) = V.

Alrighty, I got the answer. I knew that PE isn't a vector, but it didn't make sense to me (logically) that a PE in another direction is the same sign as PE in another direction. (There's a picture that came along with the problem and the moon and the earth were on opposite side of the object). It make sense on paper then since scalar quantities don't have direction but it's kinda counter-intuitive for me( still is).

9. Nov 17, 2014

### Simon Bridge

Since PE is not a vector, there is no such thing as "PE in a particular direction". This is the statement that makes no sense.

It is hard to get used to working with energy when you are used to forces - you just have to get used to it.

PE is a property of an object at a position - the PE at a particular position is the sum of the PE at the same position from all sources.
The gradient of the PE function tells you the force ... this is probably what you are intuitively thinking of as PE in one direction or another.