What Is the Minimum Mechanical Energy Needed to Change a Satellite's Orbit?

[jamesbiomed]

Homework Statement

Ok, this really IS my last question here.

Again, I have a fairly close answer that isn't right.

A 1630-kg communications satellite is released from a space shuttle to initially orbit the Earth at a radius of 9. 106 m. After being deployed, the satellite's rockets are fired to put it into a higher altitude orbit of radius 3. 107 m. What is the minimum mechanical energy supplied by the rockets to effect this change in orbit?

Homework Equations

Minimum Escape Speed:
V=sqrt(2GM/R)
G=6.67*10^-11
M=5.97*10^24

R should be the radius in whatever case. (It's the radius of the orbit so it is indeed the distance to Earth's center)


KE=1/2 m v^2

The Attempt at a Solution

I plugged both radius's into two separate eq'ns

And got

Vi=sqrt(2GM/Ri)=9406.84
Vf=sqrt(2GM/Rf)=5152.339

Then I put on my thinking cap and decided (Vi+Vf)/2 would give me the average velocity needed.

Then plugged into 1/2 mv^2 and got 4.30 e10 J

the correct answer is 2.53 e 10 J so I think my scope/basic methods are correct.

Any ideas?

Thanks a lot

 

[TSny]

You've forgotten an important type of energy.

 

[TSny]

Check this formula: V=sqrt(2GM/R)

 

[HallsofIvy]

You titled this "Escape speed" and mention escape speed in the text but this problem has nothing to do with escape speed because the object stays in orbit. There is, of course, a increase in potential energt but, if I am not mistaken, the speed necessary to orbit at a greater height is less so there will be a decrease in kinetic energy. (Ah, while I was typing this, TSny gave the formula for orbital speed.)

 

[jamesbiomed]

You mean as in E=1/2 m v^2- GMm/Re+h?

I thought kinetic meant mechanical-my fault.

So I'm going to try the average kinetic energy minus the difference in potential energy

 

[jamesbiomed]

Check this formula: V=sqrt(2GM/R)

You titled this "Escape speed" and mention escape speed in the text but this problem has nothing to do with escape speed because the object stays in orbit. There is, of course, a increase in potential energt but, if I am not mistaken, the speed necessary to orbit at a greater height is less so there will be a decrease in kinetic energy. (Ah, while I was typing this, TSny gave the formula for orbital speed.)

Sorry, didn't see either of these before I posted most recently===

TSny was telling me to check that formula, I think, because it is the one for escape speed.

Is orbital speed sqrt (GM/R)?

My book honestly doesn't have an orbital formula.

 

[TSny]

Yes, now you have the correct formula for the speed while in a circular orbit. Note that the speed will decrease in going to a larger orbit. So, the kinetic energy will decrease. But you'll also need to consider the potential energy. You don't need to write the radius as R_e + h since they give you the radius of the orbit directly.

 

[TSny]

Is orbital speed sqrt (GM/R)?

My book honestly doesn't have an orbital formula.

You can easily derive this formula from F = ma for circular motion where F is the force of gravity and a = v²/R for centripetal acceleration.

 

[jamesbiomed]

You can easily derive this formula from F = ma for circular motion where F is the force of gravity and a = v²/R for centripetal acceleration.

Oh I see, thank you

 

[jamesbiomed]

The value I'm getting for KE is very, very small. Subtracting the values for PE and dividing by two got my the correct answer on a practice

 

[jamesbiomed]

Basically, that's given me a lucky guess and it doesn't hold up. But the values I'm getting are normally about twice where they should be

 

[TSny]

Total orbital energy E = KE + PE. What is the change in total orbital energy as you go from the smaller orbit to the larger orbit? Where does the gain in energy come from?