Deriving gravitational potential energy (by bringing mass from infinity to r)

[Miraj Kayastha]

I understand the derivation of gravitational potential energy when an object is moved away from Earth but when I try to derive the formula for gpe by considering a test mass moving from infinity to r then I end up with a positive gravitational potential energy. Because integrating F.dr from infinity to r is equal to (magnitude of F) times (dr) times (cos 0) and I get a positive gravitational potential energy?

Can somebody show me the derivation of gpe considering a mass brought from infinity to r with detailed explanation (especially the signs)?

 

[xAxis]

Integrating F*dr from infinity to r gives you the work done. In the infinity the energy of system was 0. The work done gives you the change in kinetic energy. Since you got positive result, Ep + Ek = 0, therefor potential energy must be negative.
So everything is OK

 

[Miraj Kayastha]

I did not get my answer from that. I need the derivation

 

[Doc Al]

Because integrating F.dr from infinity to r is equal to (magnitude of F) times (dr) times (cos 0) and I get a positive gravitational potential energy?

What are you taking as your force F? What direction does it have?

 

[PJC01]

Sure, I can provide a detailed explanation of the derivation of gravitational potential energy when considering a test mass moving from infinity to a distance of r from a larger mass (such as the Earth).

First, let's define some variables:
- m: mass of the larger object (e.g. Earth)
- M: mass of the test object
- r: distance between the two objects
- G: gravitational constant (6.67 x 10^-11 Nm^2/kg^2)

Now, let's consider the work done by the gravitational force as the test object is moved from infinity to a distance of r from the larger object. This work is equal to the change in potential energy of the test object, and can be represented by the following equation:

W = ∆U = U(r) - U(∞)

Where U(r) is the potential energy of the test object at a distance of r from the larger object, and U(∞) is the potential energy at infinity (where the potential energy is defined as zero).

To calculate the work done, we can use the formula for work:

W = F∆r

Where F is the force acting on the test object and ∆r is the displacement of the test object.

In this case, the force acting on the test object is the gravitational force, given by Newton's Law of Gravitation:

F = GMm/r^2

Substituting this into the equation for work, we get:

W = GMm/r^2 * ∆r

Now, we can express ∆r as the difference between the final distance (r) and the initial distance (infinity):

∆r = r - ∞ = r - ∞ = r

Since the initial distance is infinity, we can assume that the test object starts at rest, and therefore its initial kinetic energy is zero. This means that all the work done by the gravitational force goes into changing the potential energy of the test object.

Substituting this into the equation for work, we get:

W = GMm/r^2 * r

= GMm/r

Since we defined the work done as the change in potential energy, we can equate this to the change in potential energy of the test object:

∆U = W = GMm/r

Now, to find the potential energy at a distance of r, we can rearrange this equation to solve for U(r):

U(r