Calculate Potential Energy between Two Masses in Space with Work Done

  • Thread starter Thread starter BillyCheung
  • Start date Start date
  • Tags Tags
    Work Work done
AI Thread Summary
The discussion focuses on calculating gravitational potential energy (P.E.) between two masses in space. It begins with a query about using the work done formula, Work = F x s, in this context. A response clarifies that the gravitational force (F) is not constant in a gravitational field, making direct application inappropriate. Instead, integration is suggested to derive the gravitational potential energy formula, E(grav) = -GmM / r. The work required to move an object from infinity to a specific point in the gravitational field is expressed through integration.
BillyCheung
Messages
9
Reaction score
0
Dear all

I know that E(grav) = -GmM / r, Can I use Work done = F x s to calculate the P.E between two mass of the space? Thank a lot.Good Bye

Billy
 
Physics news on Phys.org
Originally posted by BillyCheung
Dear all

I know that E(grav) = -GmM / r, Can I use Work done = F x s to calculate the P.E between two mass of the space? Thank a lot.Good Bye

Billy
No, I don't think so because F isn't a constant in a G-field, but we can use integration to derive the formula, E(grav) = -GmM / r.

dw = - F dx

\int ^{w'}_{0} dw = - \int ^{\infty}_{r} \frac{GMm}{x^2} dx

w' = -\frac{GMm}{r}

Where w' is the work required to take an object from infinity to a particular point in a G-field, with distant r away from the other object.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top