How to Calculate Work for Lunar Lander Orbit Change

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Homework Help Overview

The discussion revolves around calculating the work required for a lunar lander to change its orbit from 10 km to 100 km above the moon's surface. The problem involves gravitational potential energy and kinetic energy considerations in the context of orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the difference in gravitational potential energy between the two orbits and integrating the force over distance. There are attempts to compute the change in energy and questions about the calculations of orbital speeds and kinetic energy.

Discussion Status

Some participants have provided guidance on using gravitational potential energy formulas and integrating forces, while others express confusion about the calculations and the significance of the results. Multiple interpretations of the problem are being explored, particularly regarding the correct application of formulas and the assumptions about distances.

Contextual Notes

There is an emphasis on ensuring that the distance used in calculations refers to the center of the moon rather than the surface. Participants are also navigating the complexities of kinetic energy in relation to orbital speeds.

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Homework Statement


A 6000 kg lunar lander is in orbit 10 km above the surface of the moon. It needs to move out to a 100 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


U_s = -GMm/r
F_g = GMm/r^2


The Attempt at a Solution


I tried to calculate the difference in force between the two orbits and then calculate the change in energy to get the work, but I've had no luck.
 
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You would need to integrate the force over the distqance the object moves, or
use the formula for the potential energy at a a distance 'r' , pe= -GMm/r

Remember r is the total distance from the centre of the moon
 
I'm not sure how to do that.. I get:

U_g(10km) = -2.67*10^17
U_g(100km) = -2.67*10^16

the difference is only 26.7J... that can't be right.
 
sam2k2002 said:
I'm not sure how to do that.. I get:

U_g(10km) = -2.67*10^17
U_g(100km) = -2.67*10^16

the difference is only 26.7J... that can't be right.

I can't see why you think the difference is only 26.7 J. But this does not matter because:

The r in -GMm/r is the distance to the center of the moon, not the distance to the surface.

You'll need to calculate the orbital speed at 10 and 100 km as well and account for the
difference in kinetic energy.
 
How do I calculate the orbital speed?

I assume after I get that, i can compute the change in kinetic energy?
 
sam2k2002 said:
How do I calculate the orbital speed?

I assume after I get that, i can compute the change in kinetic energy?

For a circular orbit: centripetal acceleration = gravity
 
so g=v^2/r ?
 
sam2k2002 said:
so g=v^2/r ?

Yes. Of course you have g = GM/r^2 here for the acceleration of gravity
(M is mass of moon)
 

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