Calculating the Total Work Done by Gravity

AI Thread Summary
The discussion revolves around calculating the total work done by gravity on a satellite fragment after it is ejected from orbit and falls to Earth. The initial attempt at the calculation resulted in an incorrect value due to a mistake in handling orders of magnitude and unit conversions. Participants suggest using symbolic algebra to simplify the calculation process and emphasize the importance of converting the Earth's radius to meters. The original poster realizes that overlooking basic unit conversions led to their confusion and wasted time. This highlights the significance of careful unit management in physics problems.
Simon777
Messages
35
Reaction score
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.
 
Physics news on Phys.org
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;

\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)
 
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;

\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)

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
 
Convert your Earth radius to meters!
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top