Work done by gravity on falling object

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SUMMARY

The discussion focuses on calculating the work done by gravity on a 3000 kg space vehicle falling from a height of 3500 km to the Earth's surface. The key equation used is the change in gravitational potential energy, expressed as W = mgh, where g is the gravitational acceleration. The gravitational force varies with distance from the Earth's center, necessitating the use of the Law of Gravitation (GMm/r²) to accurately determine the force acting on the vehicle. The correct approach involves recognizing that the potential energy at the Earth's surface is not zero, as it is based on the distance to the Earth's center.

PREREQUISITES
  • Understanding of gravitational potential energy (mgh)
  • Familiarity with the Law of Gravitation (GMm/r²)
  • Knowledge of Newton's Second Law (\SigmaF=ma)
  • Basic calculus for setting up integrals
NEXT STEPS
  • Learn how to derive gravitational acceleration (g) at varying distances from the Earth's center
  • Study the concept of work done by variable forces in physics
  • Explore integral calculus applications in physics, specifically for calculating work
  • Investigate the implications of gravitational potential energy in orbital mechanics
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Students studying physics, particularly those focusing on mechanics and gravitational forces, as well as educators looking for examples of gravitational work calculations.

Serik
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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.
 
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Serik said:

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.
 
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.
 

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