Calculating Work to Put Spacecraft in Orbit

[AudenCalbray]

Homework Statement

How much work is done moving a 1500 kg spacecraft from the Earth's surface into a fixed orbit at a height of 300 km above the Earth's surface?

Homework Equations

Eg = -Gm1m2/r
Ek = 1\2mv^2
mass of earth: 5.98 x 10^24 kg
radius of earth: 6.38 x 10^6 m

The Attempt at a Solution

So the change in Eg plus the change in Ek is equal to 0. I solved the problem without using the 300 km or the mass of the spacecraft and I know it is wrong. I got the escape velocity (11000 m/s) and then got the kinetic energy, assuming Ek2 was 0.
I restarted the problem, and now my conservation of energy equation reads: Ek = Eg2-Eg1, again assuming Ek2 is zero, which is wrong, but I do not know how the get the change in energy otherwise. I would love some help on this. Thank you.

 

[LawrenceC]

Gravitational acceleration is a function of distance from the earth. Therefore the potential energy change is a calculus problem.

 

[AudenCalbray]

Gravitational acceleration is a function of distance from the earth. Therefore the potential energy change is a calculus problem.

Sorry, I still don't understand how to solve this.

 

[LawrenceC]

The work required equals the change in potential and kinetic energies. Both are zero on Earth's surface. You have to achieve an escape velocity at a certain altitude. The escape velocity is used to compute the kinetic energy while the mass raised to an altitude is used to compute the potential energy. The sum of the two is the work required. However, the potential energy is a calculus problem necessitating an integration. Integration is calculus which you probably have not studied.

You can avoid the integration and accept some error by choosing a mean value of gravitational acceleration and treating it as a constant. This is probably not a bad idea in view of the fact that air resistance is being ignored.

 

[Doc Al]

So the change in Eg plus the change in Ek is equal to 0.

Why do think this?

The work done will equal the change in total energy. What's the final energy of the orbiting spacecraft ? What's the initial energy?

 

[AudenCalbray]

Why do think this?

The work done will equal the change in total energy. What's the final energy of the orbiting spacecraft ? What's the initial energy?

They want the kinetic work, so the change in kinetic energy is the answer. I think.

 

[Doc Al]

What is 'kinetic work'?

 

[AudenCalbray]

What is 'kinetic work'?

Change in kinetic energy, as I understand it.

 

[Doc Al]

Change in kinetic energy, as I understand it.

Work is change in total energy.

 

[AudenCalbray]

Work is change in total energy.

If it was, wouldn't work always be 0 then?

 

[Doc Al]

If it was, wouldn't work always be 0 then?

No. Why do you think the total energy doesn't change?

 

[AudenCalbray]

No. Why do you think the total energy doesn't change?

This is a conservation of energy question.

 

[Doc Al]

This is a conservation of energy question.

It's an energy problem, but energy is not conserved. Work is being done!

Work = ΔPE + ΔKE

Don't assume that ΔPE + ΔKE = 0.

 

[AudenCalbray]

It's an energy problem, but energy is not conserved. Work is being done!

Work = ΔPE + ΔKE

Don't assume that ΔPE + ΔKE = 0.

Okay, so how do I solve it?

 

[Steely Dan]

Gravitational acceleration is a function of distance from the earth. Therefore the potential energy change is a calculus problem.

Yes, but 300 km is not very large compared to the radius of the Earth (~6000 km) so if this was done without calculus it would still give a reasonable answer.

You have to achieve an escape velocity at a certain altitude. The escape velocity is used to compute the kinetic energy while the mass raised to an altitude is used to compute the potential energy. The sum of the two is the work required. However, the potential energy is a calculus problem necessitating an integration. Integration is calculus which you probably have not studied.

It is probably confusing to use the term "escape velocity" when the Earth's gravitational field is not actually being escaped here.

Work is change in total energy.

It it always clearest to remember the work-kinetic energy theorem. Work is a change in kinetic energy; if the kinetic energy does not change, that means the net work done during the process is zero. However, in this case the kinetic energy does change (since the spacecraft starts at rest and ends at orbital speed). So there is a net amount of work being done to the spacecraft . The question is slightly ambiguous; if you are being asked how much work is done only by the thrust of the spacecraft itself, then you have to subtract out the work done by the gravitational field first.

 

[Doc Al]

Yes, but 300 km is not very large compared to the radius of the Earth (~6000 km) so if this was done without calculus it would still give a reasonable answer.

Any calculus needed has already been done. Note the formula for gravitational PE in the first post.

It it always clearest to remember the work-kinetic energy theorem. Work is a change in kinetic energy; if the kinetic energy does not change, that means the net work done during the process is zero. However, in this case the kinetic energy does change (since the spacecraft starts at rest and ends at orbital speed). So there is a net amount of work being done to the spacecraft . The question is slightly ambiguous; if you are being asked how much work is done only by the thrust of the spacecraft itself, then you have to subtract out the work done by the gravitational field first.

When they ask for the work done, I highly doubt they mean to include the work done by gravity. (That's reflected in the gravitational PE term.) They mean the work that must be done by some applied force to get the spaceship into orbit. For example: If you are asked how much work is required to raise something x meters, would you think they want an answer of zero? After all, the net work would be zero (assuming the speed doesn't change).

But I admit that it is somewhat ambiguous.

It is true that the work done by all forces (including gravity) equals the change in KE. But if you want the work done by some applied force (not including gravity), it equals the change in KE+PE.

 

[Steely Dan]

When they ask for the work done, I highly doubt they mean to include the work done by gravity. (That's reflected in the gravitational PE term.) They mean the work that must be done by some applied force to get the spaceship into orbit. For example: If you are asked how much work is required to raise something x meters, would you think they want an answer of zero? After all, the net work would be zero (assuming the speed doesn't change).

I tend to assume that the person who wrote the question meant what they said unless otherwise specified. At any rate, as long as you specify which question you are answering, you should be fine.

 

[AudenCalbray]

I tend to assume that the person who wrote the question meant what they said unless otherwise specified. At any rate, as long as you specify which question you are answering, you should be fine.

Thank you both very much. :)