Question

What is the total potential energy of earth, if it were the only mass in the universe, and has mass M=1.53.10^24 kg
and universal gravitational constant
G=6,6732.10^-11Nm^2/kg^2
and radius
r=6353.10^3m

Do I need a double integral?

Answer 1

Well, using the equation: F=G*(m1m2/r^2)
since it is the only mass remove one of them. Then that will yield you a Force. From there we can use W=Fd W being work and d being distance. If we use d as being the radius and F as being the force we solved for, then we can find the work which is also the potential energy at a maximum height. :)

Answer 2

so
\[W=\left( G.M _{earth}\over r _{earth} \right) = 62.5 MJ \]
isn't that a bit low?

Answer 3

that looks right because you dont have another mass and it should be a -GM/r

because the potential energy is the amount of energy required to move an object from infinity to a point. The potential energy is assumed to be 0 at infinity so if you move anywhere but infinity its negative.

Answer 4

ok, so potential energy is negative but still 62,5 Mega Joules to lift whole earth (that is all atoms) apart to infinity? I guess it is the wrong formula

Answer 5

yes that is correct, all i did to the formula was add a negative sign

Answer 6

i got
\[W= -7,1377.10^{31} Joule\]

Answer 7

with infinitesimal calculus.
assuming density being constant.

Answer 8

i do not understand the negative part

Answer 9

you cannot have negative energy as it is simplym a scalar thus it is just magnitude it should be positive

Answer 10

this negative energy is just a convention: Potential energy is defined as zero when two masses are apart with distance infinite. As the masses come closer to eachother, work is done by the gravity force, and energy is thus negative, since it would cost energy to get the two masses back apart towards infinite distance. You can see this in the formula of gravitation of newton, when you integrate the formula for work W= F.ds

Answer 11

dW=Fds=>
\[W = \int\limits_{\infty}^{r}F.ds = \int\limits_{\infty}^{r}m.g.ds= \int\limits_{\infty}^{r} \left( m.G.M.ds \over r^2 \right) = m.M.G \int\limits_{\infty}^{r}\left( ds \over r^2 \right)\]
result
-m.M.G /r
the work needed to get a mass from infinity to r is negative.
If you change the integration boundaries (from r to infinite) then you get positive.

Answer 12

ok i get what your saying. it is just a convention though.

Answer 13

yes, you can choose the potential energy of an object anywhere to be zero, and start from there any calculations. Is thus arbitrary number.

Answer 14

ya :P