Calculating the Total Work Done by Gravity

In summary: Using meters instead of kilometers, I get 5.65x10^7 J, which makes much more sense. In summary, the total work done by gravity on the satellite fragment is 5.65x10^7 J.
  • #1
Simon777
35
0

Homework Statement


A satellite in a circular orbit around the Earth with a radius 1.011 times the mean radius of the Earth is hit by an incoming meteorite. A large fragment (m = 83.0 kg) is ejected in the backwards direction so that it is stationary with respect to the Earth and falls directly to the ground. Its speed just before it hits the ground is 355.0 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37·10^3 km; Mass of the earth= ME 5.98·10^24 kg.


Homework Equations


Gravitational PE= (-GmM)/R

The Attempt at a Solution



Delta PE= PE intial - PE final

= (-(6.67x10^-11) (83kg) (5.98x10^24kg))/((1.011) (6.37x10^3)) - (-(6.67x10^-11) (83kg) (5.98x10^24kg))/(6.37x10^3)

= 5.65x10^10J

This is incorrect and I have tried for hours to get something to work so any help would be greatly appreciated.
 
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  • #2
Something happened to your orders of magnitude. I think you're result is about 1000x to big. Check your math.

You can save yourself a lot of digit pushing if you do some of the algebra symbolically ahead of time:

Let r = 1.011; M = Mass of Earth; R = radius of Earth; m = mass of fragment;

[tex] \Delta E = \left(\frac{G M m}{R} - \frac{G M m}{r R}\right) = \frac{G M m}{R}\left(1 - \frac{1}{r}\right) [/tex]
 
  • #3
gneill said:
Something happened to your orders of magnitude. I think you're result is about 1000x to big. Check your math.

You can save yourself a lot of digit pushing if you do some of the algebra symbolically ahead of time:

Let r = 1.011; M = Mass of Earth; R = radius of Earth; m = mass of fragment;

[tex] \Delta E = \left(\frac{G M m}{R} - \frac{G M m}{r R}\right) = \frac{G M m}{R}\left(1 - \frac{1}{r}\right) [/tex]

That does make it easier, thank you. Using it, I still end up with 5.65x10^10J. Perhaps one of my terms is wrong. This is what I used:
M=5.98·10^24 kg
G=6.67x10^-11
m=83.0kg
R= 6.37·10^3 km
r= 1.011
 
  • #4
Convert your Earth radius to meters!
 
  • #5
gneill said:
Convert your Earth radius to meters!

I can't believe I wasted all that time because I overlooked something so basic as units. Thank you so much for helping me realize this. I was over thinking it and thought I needed to factor something else in.
 

Related to Calculating the Total Work Done by Gravity

What is the formula for calculating the total work done by gravity?

The formula for calculating the total work done by gravity is: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the change in height.

What is the unit of measurement for work done by gravity?

The unit of measurement for work done by gravity is joules (J).

How is the direction of work done by gravity determined?

The direction of work done by gravity is determined by the direction of motion of the object. If the object is moving in the same direction as the force of gravity, the work done is positive. If the object is moving in the opposite direction, the work done is negative.

Can the work done by gravity be negative?

Yes, the work done by gravity can be negative if the object is moving in the opposite direction of the force of gravity. This means that the force of gravity is doing work to slow down the object's motion.

What other factors can affect the work done by gravity?

The work done by gravity can also be affected by external forces acting on the object, such as air resistance or friction. These forces can either increase or decrease the total work done by gravity.

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