• Support PF! Buy your school textbooks, materials and every day products Here!

Work done by gravity on falling object

  • Thread starter Serik
  • Start date
  • #1
8
0

Homework Statement



A 3000 kg space vehicle, initially at rest, falls vertically from a height of 3500 km above the Earth's surface.

Determine how much work is done by the force of gravity in bringing the vehicle to the Earth's surface.

Homework Equations



Gravitational potential energy: mgh
Law of Gravitation: GMm/r2
Newton's Second Law: [tex]\Sigma[/tex]F=ma


The Attempt at a Solution



The work done should be the change in potential gravitational energy, correct?

There's zero gravitational potential energy when the spacecraft is on earth (y=0). When the spacecraft's at the height above earth from which it will fall, its gravitational potential energy is:

mcraft*g*h

Where g is the gravitational acceleration. This can be found with:

Fg=ma
a=Fg/mcraft

Where Fg = GMearthmcraft/h2

Putting it altogether:

W=mcraft*[(GMearthmcraft/h2)/mcraft]*h - 0

Am I doing this correctly? Also, since we haven't technically covered potential energy, how would I set up an integral to find out how much work was done, since the force of gravity (and hence its acceleration) varies with r, the distance to the center of the earth?

Thank you for your help.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4

Homework Statement



A 3000 kg space vehicle, initially at rest, falls vertically from a height of 3500 km above the Earth's surface.

Determine how much work is done by the force of gravity in bringing the vehicle to the Earth's surface.

Homework Equations



Gravitational potential energy: mgh
Law of Gravitation: GMm/r2
Newton's Second Law: [tex]\Sigma[/tex]F=ma


The Attempt at a Solution



The work done should be the change in potential gravitational energy, correct?

There's zero gravitational potential energy when the spacecraft is on earth (y=0). When the spacecraft's at the height above earth from which it will fall, its gravitational potential energy is:

mcraft*g*h

Where g is the gravitational acceleration. This can be found with:

Fg=ma
a=Fg/mcraft

Where Fg = GMearthmcraft/h2

Putting it altogether:

W=mcraft*[(GMearthmcraft/h2)/mcraft]*h - 0

Am I doing this correctly? Also, since we haven't technically covered potential energy, how would I set up an integral to find out how much work was done, since the force of gravity (and hence its acceleration) varies with r, the distance to the center of the earth?

Thank you for your help.
So long as you realize that the h you're talking about is the distance to the center of the Earth.
And the distance from the surface to the center becomes your rest potential.
 
  • #3
8
0
Ahhh, thanks.

In the corner of my notes, I scrawled h=Habove earth + Rearth, but failed to carry that substitution through the remainder of my calculations.

So at the surface, which I shouldn't really call y=0, the spacecraft will have gravitational potential energy because it's a distance Rearth from the center of the planet.
 

Related Threads for: Work done by gravity on falling object

Replies
10
Views
66K
Replies
7
Views
3K
Replies
4
Views
3K
  • Last Post
Replies
15
Views
9K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
3K
Top