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Well here I am again, with another question because the people in my class are retarded and cant help me on our class forums lol…

So here it is:

Energy is required to move a 1090 kg mass from the Earth's surface to an altitude 2.01 times the Earth's radius (6.37 x 10^6 meters). What amount of energy is required to accomplish this move? Hint: G and Me can be eliminated in favor of parameter of the gravitational acceleration at the surface of the earth; i.e., the acceleration of gravity on Earth is 9.8 m/s^2. Answer in units of J.

Any help on this would be appreciated since I have tried 4 times already and have 3 guesses left :eek:

Thanks in advance!

oh man!

this is at the top of my head!

if only my AP Physics teacher emphasized more on the equations rather than the "concepts"…. :doh:

From: legendaryhero90
oh man!

this is at the top of my head!

if only my AP Physics teacher emphasized more on the equations rather than the "concepts"…. :doh:

Well lemme know if it comes to you then :D

Working on it. Will edit in a little with the answer. (If I can figure it out)

Edit:

Actually we learned this at the beginning of the year.

mgh = PEg (mass x gravity x height = Potential Energy of gravity)

so
1090kg x 9.8m/s x 12200700meters = approximately 1.3 x 10^11 J.

Your welcome.

Work = the integral of force over a distance. F = G*m1*m2/r^2 where G is the gravitational constant. So basicly the only thing that is changing here is the distance you are from the earth.

so, (I wish I could do symbols) Work (energy) needed = G * m1 * m2 * integral from radius of the earth to 2.01 x the radius of the earth of dr / 1 / r^2 which gives us Energy = (-(G * m1 * m2) / r) from radius of the earth to the outer shell. Giving you Energy = (-G * m1 * m2) / (2.01 * radius of the earth) + (G * m1 * m2) / radius of the earth.

Of course, G is the gravitational constant, m1 is the mass of the earth, m2 is the mass of your object. Make sure you plug in SI units (kg, meters, ect) Don't know what the hint is all about, but this should work (let me plug it into my calculator to give you an exact result)

From: mickaliscious
Working on it. Will edit in a little with the answer. (If I can figure it out)

Edit:

Actually we learned this at the beginning of the year.

mgh = PEg (mass x gravity x height = Potential Energy of gravity)

so
1090kg x 9.8m/s x 12200700meters = approximately 1.3 x 10^11 J.

Your welcome.

This answer works if you are close to the surface of the earth, however, you are not (gravity will vary as you move farther out) my answer will be more exact.

I get 6.89047244 × 10^10 J Sig. Figs puts it at 6.9 x 10^10 The formula I used is correct but I may have entered some numbers in wrong, so you'll have to check that.

*edit, helps if you give the calculator the correct numbers*

From: boberman
This answer works if you are close to the surface of the earth, however, you are not (gravity will vary as you move farther out) my answer will be more exact.

I get 1.37120402 × 10^15 Sig. Figs puts it at 1.4 x 10^15 The formula I used is correct but I may have entered some numbers in wrong, you you'll have to check that.

Yeah, you're right. I was assuming gravity was constant. I have no idea how to solve it otherwise. But theres no way that as gravity decreases you would get a higher amount of energy to move it (i.e. my number 10^11 is smaller than 10^15)

From: mickaliscious
Yeah, you're right. I was assuming gravity was constant. I have no idea how to solve it otherwise.

The equation I used above will do it, You'll need to use Calculus to solve this problem though. (not hard calculus, just calculus). So it is the integral of the change of force over a distance that gives you work done.

To get the force as a function of distance look up Gravitational Constant in the wikipedia. They tell you all you need to know. :)

Oh, yeah about that, I was unsure that the difference would be that big either, I got my result by plugging the numbers into google (couldn't find my TI-86) so your answer could very well be in the right magnitude. But Im very certain mine will be more accurate (with correct numbers of course :D)

:) Found it, It is actually quite a big difference, (my problem was that I did a 10^2 instead of 10^6 in one of my terms) so the REAL answer is 6.89047244 × 10^10 J *Edits first answer)

thanks for the help guys, but i said screw it at about 11:00 and just went to bed =/