What Is the Work Done When Moving a Mass from X to Y?

In summary: The property that gravitational field is conservative means that the work done by the gravitational field on an object is equal to the object's change in potential energy. This is why you can use the potential energy at 2R to find the work done by the field on the mass at 2R.
  • #1
DumbNut
3
0

Homework Statement


X & Y are two points at respective distances R and 2R from the centre of the Earth, where R is greater than the radius of the Earth. The gravitational potential at X is -800kJkg^-1. When a 1kg mass is taken from X to Y, what is the work done on the mass?

Homework Equations


Gravitational Potential = -GM/r
U= -GMm/r

The Attempt at a Solution


I have thought over this question and tried using integration to make sense of it since gravitational potential and U varies w/ r. But when I tried thinking along the lines of integration, I find myself wondering what does the integrated value of graph U against r represent as it does not concur with what I am trying to find with regards to the concept of homogeneity. Is my idea of trying to integrate wrong? Or is my concept wrong to begin with? Please help, thank you.
 
Last edited:
Physics news on Phys.org
  • #2
Welcome to PF,

I don't think you need to integrate. Since you know the gravitational potential (often called [itex]\psi[/itex] or [itex] \phi[/itex]) at R, you can easily figure out what it is at 2R, since the form of the function [itex]\phi(r)[/itex] is known.

Once you get the potential at 2R, you know the potential energy that this particular mass therefore has at 2R. You can compare that to the potential energy it had back when it was only at R. Recall that the work done by a conservative force on an object is equal to the negative of that object's change in potential energy.
 
  • #3
Thank you! I guess I thought about it too mathematically, now it makes sense to me.
 
  • #4
The work done in moving a mass between two points is the difference in the potential energy (U). As far as I know, integrating U doesn't have any physical significance. By definition, U is the work done by the gravitational field of an object to move it from a point to infinity. Thus if you consider the difference in the potential energies at any two points, you can find the work required to move a body between those two points.

EDIT: Looks like someone else beat me to it!
 
  • #5
Sorry, but I can't understand your question.
However, why don't you use the property that gravitational field is a conservative one?
 
  • #6
Yeap, I tried that and it worked :D thanks guys.
 

FAQ: What Is the Work Done When Moving a Mass from X to Y?

What is the relationship between distance and mass in terms of work done?

The relationship between distance and mass in terms of work done is that the amount of work done is directly proportional to the distance traveled and the mass moved. In other words, the greater the distance and mass, the more work is required to be done.

How does distance affect the amount of work done?

Distance affects the amount of work done because work is defined as the product of force and distance. This means that the greater the distance traveled, the more work is required to be done.

What is the formula for calculating work done?

The formula for calculating work done is W = F x d, where W is the work done, F is the force applied, and d is the distance traveled.

How does mass play a role in the calculation of work done?

Mass plays a role in the calculation of work done because it is a measure of an object's resistance to acceleration. The greater the mass, the more force is required to move it a certain distance, resulting in more work being done.

Can distance and mass affect the efficiency of work done?

Yes, distance and mass can affect the efficiency of work done. The more distance and mass involved, the lower the efficiency will be as more energy is required to overcome the resistance. This is why it is important to consider distance and mass when trying to increase efficiency in work processes.

Back
Top