Is it true to define

1. Gravitational potential - work done per unit mass (AGAINST THE GRAVITATIONAL FORCE) when moving an object from infinity to a given point

so that we can conclude that it will be negative as when we are moving from infinity to a point, we are not doing work against gravity. gravity is doing the work but not we so that's why it is negative work done per unit mass

According to the same logic:

GPE is the work done against the gravity to bring a mass to a given point in space from infinity and the same explanation(no work is done against gravity, the work is done in the direction of gravity so it's a negative work and hence GPE has minus sign in its formula)

1. Gravitational potential - work done per unit mass (AGAINST THE GRAVITATIONAL FORCE) when moving an object from infinity to a given point

so that we can conclude that it will be negative as when we are moving from infinity to a point, we are not doing work against gravity. gravity is doing the work but not we so that's why it is negative work done per unit mass

According to the same logic:

GPE is the work done against the gravity to bring a mass to a given point in space from infinity and the same explanation(no work is done against gravity, the work is done in the direction of gravity so it's a negative work and hence GPE has minus sign in its formula)

I like to remember that force is the negative derivative of potential- if our function of force is F(x) and our potential is V(x), then -d[v(x)]/dx=F(x). Try it out on the equation for GPE with spheres, V(x)=-GM/x. You'll find you're correct.

Now in response to your question, if you were to integrate force with respect to displacement as displacement decreases, you would find your potential energy has decreased.

As shown below:

V(x)=∫^{R}∞ F(x) dx =∫^{R}∞ GMm/(x^{2})dx = [-gmm/(x)]^{R}∞

Where x is distance between the two centres of mass. As you can see, potential energy in the infinity part of the integral approaches zero.

I think that this should prove that you are correct! If I haven't made any mistakes.

My infinity signs were meant to be at the bottom of each integral by the way (idk how to do math on here yet). Try sketching the graph for force against displacement to check! ^^

Now in response to your question, if you were to integrate force with respect to displacement as displacement decreases, you would find your potential energy has decreased.

As shown below:

V(x)=∫

Where x is distance between the two centres of mass. As you can see, potential energy in the infinity part of the integral approaches zero.

I think that this should prove that you are correct! If I haven't made any mistakes.

My infinity signs were meant to be at the bottom of each integral by the way (idk how to do math on here yet). Try sketching the graph for force against displacement to check! ^^

(edited 1 month ago)

Original post by Hep5

I like to remember that force is the negative derivative of potential- if our function of force is F(x) and our potential is V(x), then -d[v(x)]/dx=F(x). Try it out on the equation for GPE with spheres, V(x)=-GM/x. You'll find you're correct.

Now in response to your question, if you were to integrate force with respect to displacement as displacement decreases, you would find your potential energy has decreased.

As shown below:

V(x)=∫^{R}∞ F(x) dx =∫^{R}∞ GMm/(x^{2})dx = [-gmm/(x)]^{R}∞

Where x is distance between the two centres of mass. As you can see, potential energy in the infinity part of the integral approaches zero.

I think that this should prove that you are correct! If I haven't made any mistakes.

My infinity signs were meant to be at the bottom of each integral by the way (idk how to do math on here yet). Try sketching the graph for force against displacement to check! ^^

Now in response to your question, if you were to integrate force with respect to displacement as displacement decreases, you would find your potential energy has decreased.

As shown below:

V(x)=∫

Where x is distance between the two centres of mass. As you can see, potential energy in the infinity part of the integral approaches zero.

I think that this should prove that you are correct! If I haven't made any mistakes.

My infinity signs were meant to be at the bottom of each integral by the way (idk how to do math on here yet). Try sketching the graph for force against displacement to check! ^^

thank you! that's a very clear method. I will try to derive it on my own

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