Question about Universal Gravitation

[motionman04]

How much energy is required to move a 1070 kg object from the Earth's surface to an altitude four times the Earth's radius?

Having problems with this question.

I figure it would be GMm(1/r(final) - 1/r(initial)), with the final being 4x Earth's radius, and initial being Earth's radius. I plugged in the constants for G, and M(earth), but the answer I'm getting is wrong. Please help.

 

[Sirus]

W=\Delta{E_{P_{g}}}

E_{P_{g}}=-\frac{GMm}{r}

Watch the signs.

 

[motionman04]

hmm, I tried with the signs, however, looks like there's something up with the number I'm getting

 

[vsage]

What answer are you getting? The number should be positive, and the formula you're using will fetch you a negative number.

 

[motionman04]

Well, I used (6.37x10^6)(4), to get Rfinal, since that is the Earth's radius times 4, and I use 6.37E6 for Rinitial, since the particle starts on the Earth's surface. Of course, G and M are constants, and I used 1070 for the mass of the object. Plugging everything in, I get

- (6.67E-11)(5.98E24)(1070)(1/25480000 - 1/6370000) = 5.025E10

 

[Curious3141]

You're misreading the problem : altitude of 4X the Earth's radius. They did not say to a distance from the Earth's center of mass of 4X the Earth's radius. Do you see now ?

 

[motionman04]

I see what you mean, so what would Rfinal be?

Its 2AM, the reason why I'm asking stupid questions

 

[Curious3141]

I see what you mean, so what would Rfinal be?

Its 2AM, the reason why I'm asking stupid questions

Five times the Earth's radius. If you can't see why this is so, please post.

 

[dextercioby]

Just a piece of advice:it's not indicated to (try to ) solve physics problems at 2AM,especially in the weekend,when you have enough time (i think) during the day...

Daniel.

 

[motionman04]

Five times the Earth's radius. If you can't see why this is so, please post.

Wait, but I did set that as Rfinal, four times the Earth's radius, and came out with the answer that I got before. Am I supposed to do it based on 5x the radius? If that is so, I'm kinda confused

 

[dextercioby]

Yes."move a 1070 kg object from the Earth's surface to an altitude four times the Earth's radius".I think the word "altitude" is referred wrt the Earth's surface,therefore the distance between the orbit and the Earth's center is the altitude+Earth's radius which is 5 times the Earth's radius...So 5R_{E} is your R_{fin}.

Daniel.

 

[ramin86]

Yes, that worked out. Thanks a lot for the help. I actually have one more question:

At the Earth's surface a projectile is launched straight up at a speed of 10.3 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

Now this question is listed under Universal gravitation, but looking through the formula's I can't seem to find one that can fit with this problem.

 

[dextercioby]

Apply the law of conservation of mechanical energy.Again,pay attention with the signs,as the PE is negative (by convention,it's chosen 0 at infinity).

Daniel.