Cost of setting up a sattelite?

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Homework Help Overview

The discussion revolves around the energy requirements for deploying a 100kg experimental package into orbit around the Earth, specifically focusing on the work needed to elevate the package to a height of 1,000 km and the additional work required for circular orbit stabilization.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between work and energy, questioning the appropriateness of using constant force equations versus gravitational force equations. There is discussion about the relevance of kinetic and potential energy in the context of the problem.

Discussion Status

The conversation is ongoing, with participants providing hints and questioning each other's reasoning. Some guidance has been offered regarding the use of gravitational force and the need to consider changes in potential energy, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are navigating the complexities of gravitational forces and energy calculations, with some indicating confusion over the assumptions related to uniform gravity and the application of various energy equations. There is an emphasis on careful consideration of the problem setup.

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Homework Statement



A scientist wants to put an 100kg experimental package in orbit around the Earth. The cost of deployment depends on the amount of extra energy it takes to get it into the required position i.e. how much more energy is used than just sending the rocket up there.
a) Determine the amount of work that must be done to get the package to 1 000km above the Earth’s surface.
b) Determine the amount of extra work needed to put the package into a circular orbit at this height.

Homework Equations



W=F*d
F=(G*m1m2)/r2

The Attempt at a Solution



a) W=F*d
F=FG
=(Gm1m2)r2
=(6.67 * 10-11 * 100 * ME)/(RE + 1000000)2
= 731 N
W= F*d = 730 * 1000000
= 7.31 *108
 
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Zaros said:
W=F*d
That is true only for a constant force.

You need to use something else. Hint: What is the relation between work and energy?
 
Work is equal to kinetic energy i.e. W= 1/2 mv2
Is that what your getting at?
 
What about potential energy?
 
W = KE + PE = 1/2 mv2 + mgh
Is that right?
 
No. That mgh term is assuming a uniform gravity field.
 
then should i use the F= (G*m1*m2)/r2 instead so that would lead to
F= 1/2 mv2 + h((G*m1*m2)/r2)

Now is this correct?
 
Zaros said:
then should i use the F= (G*m1*m2)/r2 instead so that would lead to
F= 1/2 mv2 + h((G*m1*m2)/r2)

Now is this correct?

Stop just throwing equations at the problem. You equated F, Force, with an energy. Not good. You need to stop and think about the problem.

What would your change in potential energy be if you were to take the package on the ground with 0 kinetic energy and put it up into space at the height required and still not let it have any kinetic energy?
 

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