How to Estimate Work Done from a Graph?

[Jimmy87]

Homework Statement

Hi,

I am stuck with the following question:

The graph (attached) shows how the Moon's gravitational pull F on a lunar lander varies with the distance 'h' from the Moon's surface. Estimate the work done by the pull of the Moon on the lunar lander as it approaches the lunar surface from a height of 200km. Explain how you made your estimate.

Homework Equations

Work done = force x distance

The Attempt at a Solution

I know that work = force x distance so I was thinking of the calculating the area under the graph but I don't think this will work as this would involve multiplying 0.88kN by 200m x 0.5 and that just can't be right because the average force has got to be somewhere between 3.6 and 2.92kN. Why doesn't this work though as I thought that finding the area under a graph involves multiplying the y-axis by the x-axis which in this case would be force x distance which is work?

 

[mfb]

You have to consider that the area should go down to zero, while your graph does not. Imagine an extension of the graph down to 0.

 

[Jimmy87]

You have to consider that the area should go down to zero, while your graph does not. Imagine an extension of the graph down to 0.

Thanks. I still don't understand though. Extension on what axis? The force at 200km is about 2.92kN and at the surface (0m) is 3.6kN so how can anything go down to zero?

 

[Jimmy87]

You have to consider that the area should go down to zero, while your graph does not. Imagine an extension of the graph down to 0.

Actually I think I see what you mean now. The force at 200m away is 2.92kN so if the force remained at this value for the 200km then you would just multiply these two numbers but since it increase do you add this value to the value underneath the curve which would be 0.88kN x 200 x 0.5. So the total sum would be:

(2.92kN x 200) + (0.88kN x 200 x 0.5) = 584 + 88 = 672kJ (I think these are the correct units)

Is this the correct answer?

 

[mfb]

Kilometers, not meters.
Apart from that: yes, that works.

The curve is not completely linear so you overestimate the work a bit.