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A 1 500kg satellite is in orbit 250km above earth's surface.

  • Thread starter jakeginobi
  • Start date
1. Homework Statement
A 1 500kg satellite is in orbit 250km above earth's surface. What minimum additional energy is needed to place this satellite in a new stable orbit 800km above earth's surface?

2. Homework Equations
W = delta Ep = -GMm(1/r2-1/r1)

Ep = -GMm/r
Ek = GMm/2r or 1/2mv^2
3. The Attempt at a Solution
I tried using the work formula = -GMm (1/2r2-1/2r1, but I get the wrong answer. The answer is 3.41 x 10^9J. I thought to change the radius of orbit from one to another is the change in potential energy which is work?
 

TSny

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The first part, W = delta Ep, is OK.
But, delta Ep = -GMm(1/r2-1/r1) is only true if there is no change in kinetic energy.
whenever an object radius of orbit is changed, there's a change in kinetic energy as well?

What if the object isn't in orbit, but a change in altitude? Would there be a change in kinetic too?
 

TSny

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whenever an object radius of orbit is changed, there's a change in kinetic energy as well?
Yes.

What if the object isn't in orbit, but a change in altitude? Would there be a change in kinetic too?
That depends on how you change the altitude.
 
Yes.

That depends on how you change the altitude.
For instance, in this question "An unpowered 1600kg object has an upward velocity of 7.0x10^3 m/s at an altitude of 100km above the earth. The object reaches a maximum altitude of 400km" would there be a kinetic energy change?
 

TSny

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3. The Attempt at a Solution
I tried using the work formula = -GMm (1/2r2-1/2r1, but I get the wrong answer.
Actually, this formula does look like it should give you the correct answer if the expression is written as -GMm (1/(2r2)-1/(2r1)). I haven't tried plugging in numbers.

Note that the total energy is E = Ek + Ep. You stated that you can write Ek as GMm/(2r), although you left out the parentheses. Thus, you should be able to show that E = -GMm/(2r).
 

TSny

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For instance, in this question "An unpowered 1600kg object has an upward velocity of 7.0x10^3 m/s at an altitude of 100km above the earth. The object reaches a maximum altitude of 400km" would there be a kinetic energy change?
Yes. There is a change in kinetic energy if there is a change in speed. What is the speed at maximum altitude?
 
Yes. There is a change in kinetic energy if there is a change in speed. What is the speed at maximum altitude?
The speed at maximum altitude is zero, but I'm not sure why it is zero
 

TSny

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However, you need to be careful with questions that ask you to calculate work. In particular, there is a difference between asking how much work you would have to do as compared to how much work the force of gravity does.

The work done by the force of gravity is always equal to the negative of the change in potential energy only. (Any change in kinetic energy is irrelevant.)
 

TSny

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The speed at maximum altitude is zero, but I'm not sure why it is zero
If the velocity is not zero, then the object is moving either upward or downward. Either way, the object cannot be at its maximum height. I'm assuming that the object is moving along a vertical line.
 

TSny

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A fairly common error in these types of problems is to confuse "altitude of the orbit" with "radius of the orbit".

I get the correct answer using your formula given in post # 1. If you show your calculation with the numbers you used, we can see if there is a mistake in how you are using the formula.
 
A fairly common error in these types of problems is to confuse "altitude of the orbit" with "radius of the orbit".

I get the correct answer using your formula given in post # 1. If you show your calculation with the numbers you used, we can see if there is a mistake in how you are using the formula.
I used -GMm(1/r - 1/r) so it's -6.67x10^-11(5.98x10^24kg)(1500kg)(1/7.18x10^6m - 1/6.63x10^6m)
 

TSny

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I used -GMm(1/r - 1/r) so it's -6.67x10^-11(5.98x10^24kg)(1500kg)(1/7.18x10^6m - 1/6.63x10^6m)
OK. This is the wrong formula. In your first post you wrote
3. The Attempt at a Solution
I tried using the work formula = -GMm (1/2r2-1/2r1,
which looks like the correct formula (except for some missing parentheses).

The question asks for the change in energy. This refers to the change in total energy, not just the change in potential energy.
 
OK. This is the wrong formula. In your first post you wrote which looks like the correct formula (except for some missing parentheses).

The question asks for the change in energy. This refers to the change in total energy, not just the change in potential energy.
Oh alright thank you, I have one more question. Is total energy always conserved if you change altitudes if friction, or heat is neglected?
 

TSny

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I have one more question. Is total energy always if you change altitudes conserved if friction, or heat is neglected?
I would need to have more information.

For example, if the object is changing altitude because it is in an elliptical orbit and there are no forces other than gravity acting, then the total energy (Ek + Ep) is conserved.

But if the altitude is changed by some external agent (as in your exercise), then the work done by the external agent will cause the total energy of the object to change.
 

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