Lunar lander in orbit around the moon and how much work the thrusters do

[kerbyjonsonjr]

Homework Statement

A 6000 kg lunar lander is in orbit 80 km above the surface of the moon. It needs to move out to a 200 km -high orbit in order to link up with the mother ship that will take the astronauts home.

How much work must the thrusters do?

Homework Equations

v=$$\sqrt{}GM/r$$
W=$$\Delta$$KE
G=6.67 x10^-11
Radius of moon= 1,737.4 km
Mass of moon=7.36 x10²²

The Attempt at a Solution

Since W=change in KE I figured I would just find the velocity before and after then plug it in. I used v=sqrt GM/r so sqrt 6.67 x10^-22* 7.36 x10²²/(1737.4+80) and got 51,972.9 m/s and then I did that again with the radius being 1,737.4+200 and got v= 50,337.7 m/s

Then I solved for change in KE=1/2mv² so 1/2(6000)(50377.7²) -1/2(6000)(51972.9²) and got -5.02 x10¹¹

I am not sure where I messed up but I know my answer is wrong. Any help would really be great.

 

[D H]

Think about what that negative value means. If it were true, things would not stay in orbit. They would instead spiral outward.

You need to calculate the change in total energy, not just the change in kinetic energy.

 

[kerbyjonsonjr]

You need to calculate the change in total energy, not just the change in kinetic energy.

Alright, so would I solve for change in E by saying KE_f + U_f -(KE_i + U_i) ? Is U equal to GMm/r?

 

[D H]

Is U equal to GMm/r?

Wrong sign.

 

[kerbyjonsonjr]

Wrong sign.

Is the rest of the setup though because I am still getting an incorrect answer

I have KE_f-GMm/r +(KE₂ - GMm/r)

So I put in 1/2(6000)(50,337²) - 6.67 x10^-11(7.36 x10^22)(6000)/(1937)= -7.6 x10^12

Then the other half is 1/2(6000)(51973²) -6.67 x10^-11(7.36 x10^22)(6000)/(1817)= -8.1 x10^12

So then that becomes -7.6 x10^12 + 8.1 x10^12 = 5 x10^11

Where did I go wrong?

 

[D H]

Units! Don't just plug in numbers. You have to make sure that the terms in your expression have consistent units.

 

[kerbyjonsonjr]

Units! Don't just plug in numbers. You have to make sure that the terms in your expression have consistent units.

Would it be the radius that needs to be in meters? Or is it something else?

 

[D H]

Yes, the radius needs to be in meters.

 

[kerbyjonsonjr]

It would seem as though I am still doing something wrong because I am now getting a negative value again.

I plugged in the value for r in meters my change in energy ended up being
7.58 x10^12 - 8.07 x10^12 which is a negative number. The only thing I did different was plug in the new r values.

 

[D H]

Show your work, with units, not just numbers.

 

[kerbyjonsonjr]

The final energy: 1/2(6000 kg)(50377² m/s) - 6.67 x 10^-11 (7.36 x10^22 kg)(6000 kg)/(1937000 m) = 7.6 x10^12 - 1.52 x10^10= 7.59 x 10^12

The initial energy: 1/2(6000 kg)(51973² m/s) -6.67 x10^-11 (7.36 x10^22 kg)(6000 kg)/(1817000m)= 8.1 x10^12 - 1.62 x10^10 = 8.08 x10^12

Then I would subtract 7.59 x10^12 - 8.08 x10^12= -4.9 x10^11. I am not sure if I am putting something in incorrectly on my calculator or if I have an error with my units.

 

[D H]

Your velocities are wrong. Show your work, and once again watch your units.

 

[kerbyjonsonjr]

I finally got it. I forgot to use m for the radius when I calculated the velocities. I redid it and finally got the correct answer of 5.01 x10^8 J. Thank you so much for taking the trouble to help me. I really appreciate it.

 

[D H]

Thanks.

Lesson learned, I hope: Don't just throw numbers around. Always work with units.