Work done by gravity on falling object

AI Thread Summary
The discussion focuses on calculating the work done by gravity on a 3000 kg space vehicle falling from a height of 3500 km. The work done is identified as the change in gravitational potential energy, which is expressed as mgh, where g is the gravitational acceleration. The participant notes that gravitational potential energy is zero at the Earth's surface and attempts to derive the work done using gravitational force equations. There is a clarification that the height should be measured from the center of the Earth, not the surface, and the participant realizes the need to adjust their calculations accordingly. The importance of integrating to account for varying gravitational force with distance from the Earth's center is also highlighted.
Serik
Messages
7
Reaction score
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: \SigmaF=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.
 
Physics news on Phys.org
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: \SigmaF=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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top