Falling Meteorite - Gravitation PE + KE

[Faint]

Homework Statement

A satellite in a circular orbit around the Earth with a radius 1.021 times the mean radius of the Earth is hit by an incoming meteorite. A large fragment (m = 65 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 369 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37 x 10^3 km; Mass of the earth= ME 5.98 x 10^24 kg.

Hint: The work done by gravity must equal the change of potential energy of the satellite fragment. Use the general formula for Gravitational Potential Energy, NOTe: PE = mgh, is only valid for small distances above the surface.

Homework Equations

PE = (-G*ME*m)/(r)
KE = 1/2 * m * v^2

The Attempt at a Solution

PEinitial + KEinitial = PEfinal +KEfinal - W

KEi = 0
PEf = 0

W = KEf - PEi

Work Done = (1/2 * (65) * (369)^2) - (-G*(5.98*10^24)*(65))/(1.021*6.37*10^3)
Work Done = 4425232.5 - (-3.9881408782 * 10^-9)
Work Done = 4425232.5 J

Not sure where I am going wrong for this problem.

 

[tiny-tim]

Hi Faint!

(try using the X² and X₂ icons just above the Reply box )

hmm … you're very confused about potential energy and work (and the question isn't helping, by giving too much information)

PEinitial + KEinitial = PEfinal +KEfinal - W
…
W = KEf - PEi

No, PE_i + KE_i = PE_f + KE_f (or ∆KE + ∆PE = 0)

and W = -∆PE …

PE is defined as (minus) the work done by a conservative force (such as gravity).

(so PE can't be defined for a non-conservative force such as friction)

You can calculate W either as -∆PE or (in this case) as ∆KE (since for some reason they've given you the final speed )

(you're probably being confused by problems such as the work done by friction on a rough slope … in that case, friction and gravity are, illogically, treated differently: friction as a force and gravity as a source of energy )

 

[D H]

You can calculate W either as -∆PE or (in this case) as ∆KE (since for some reason they've given you the final speed )

I wouldn't use the ∆KE approach to answer this question.

I suspect the very next part of the question (omitted by the OP) is something along the lines of "Explain why your calculated value for work done by gravity differs from work calculated as change in kinetic energy".

 

[Faint]

You can calculate W either as -∆PE or (in this case) as ∆KE (since for some reason they've given you the final speed )

Thank you for your help so far. Another question though, if I calculate -∆PE and ∆KE I get the following:
-PE∆ = - (-G*(5.98*10^24)*(65))/(1.021*6.37*10³)
-PE∆ = 3.9881408782 * 10⁹ J

∆KE = (1/2 * 65 * 369²)
∆KE = 4425232.5 J

These are two very different answers, which leads me to believe that I am still missing something?

I wouldn't use the ∆KE approach to answer this question.

I suspect the very next part of the question (omitted by the OP) is something along the lines of "Explain why your calculated value for work done by gravity differs from work calculated as change in kinetic energy".

This is the only part of this question I have.

 

[tiny-tim]

Hi Faint!

These are two very different answers, which leads me to believe that I am still missing something?

No, as D H hinted at, they will give different answers.

One answer is ∆KE, and the other is -∆PE …

they will only be the same if ∆KE + ∆PE = 0, in other words if (mechanical) energy is conserved.

But this fragment had to fall through the atmosphere (not mentioned in the question!), so it lost some mechanical energy (and heated up instead).

So the ∆PE is only the work done by gravity, while the ∆KE is the total work done (including the work done by friction).

(And the question only asked for the work done by gravity … the final speed seems to be a red herring )​

 

[D H]

Thank you for your help so far. Another question though, if I calculate -∆PE and ∆KE I get the following:
-PE∆ = - (-G*(5.98*10^24)*(65))/(1.021*6.37*10³)
-PE∆ = 3.9881408782 * 10⁹ J

∆KE = (1/2 * 65 * 369²)
∆KE = 4425232.5 J

These are two very different answers, which leads me to believe that I am still missing something?

Yes, you are missing something. You are not calculating ∆PE. You are missing the potential energy at the surface of the Earth.

 

[tiny-tim]

oops! I missed that!

 

[Faint]

Once again, thank you for the help. So would I be correct in saying:

-PE∆ = - (-G*(5.98*10^24)*(65))/(1.021*6.37*10³) - (-G*(5.98*10^24)*(65))/(6.37*10³)
-PE∆ = -(-3.9881408782 * 10¹² - (-4.07189183673*10¹²))
-PE∆ = -83750958444.1 J
So...
W = 83750958444.1 J ?

 

[D H]

Still off, by several orders of magnitude. Look at your units.

 

[Faint]

Forgot to multiply the radius by 1000 to convert to meters. Got it now, thanks!