# Gravitational Potential & Gravitational Potential Energy

• Jimmy87
In summary, this increases the work done to move the mass from infinity to a point in a gravitational field.

#### Jimmy87

Homework Statement
Understanding the Definitions and Model
Relevant Equations
V = - GM/R and E = - GMm/R
Hi,

I am confused about the negative aspect of these quantities. The definition in my book for gravitational potential is:

"The work done to move a unit mass from infinity to a point in a gravitational field"

I understand that the work done is negative because gravity is doing the work if you starting point is infinity. However, my book has a key points section and says it is "important to understand that moving towards the gravitating mass is a decrease in gravitational potential and moving away represent an increase". This does not make sense in terms of the definition given. If you move further away from Earth then LESS work is done per unit mass bringing it from infinity. I understand the value of the potential will be less negative/more positive so that is a larger number but the fact it increases as you get further away does not make sense with the definition of potential otherwise we would be saying that this increases:

"The work done to move a unit mass from infinity to a point in a gravitational field"

This is the definition of gravitational potential in my book and on numerous websites and it does not increase as you get further from the Earth! There is not an increase in the work done per unit mass from infinity! If it's because the number is more positive that's fine but then the definition needs changing surely!

Can someone please kindly explain. Thank you.

Jimmy87 said:
"important to understand that moving towards the gravitating mass is a decrease in gravitational potential and moving away represent an increase"
Decreases in the sense of more negative. There's no contradiction. It does not say the magnitude decreases.

• Jimmy87
Jimmy87 said:
Can someone please kindly explain. Thank you.
One idea is to use positive work. Start at some reference radius ##R_0## and calculate the (positive) work needed to move a unit mass out to a new radius ##R > R_0##. With this reference we have:
$$V^*(R) = \int_{R_0}^R \frac{GM}{r^2} \ dr = \big [-\frac{GM}{r} \big ]_{R_0}^R = - \frac{GM}{R} + \frac{GM}{R_0}$$
Now, we simply change our reference so that ##V(\infty) = 0## by subtracting the constant term ##\frac{GM}{R_0}## to give:
$$V(R) = - \frac{GM}{R}$$
Note that ##V(R)## is increasing as ##R## increases - and has a maximum of ##0## as ##R## goes to infinity.

It's perhaps a good exercise to work all this out for yourself using negative work and a mass coming in from infinity.

• Jimmy87