Gravitational formula help

1. Jan 31, 2010

aloshi

hi!
in my book they are trying to derive a formula for how much energy is needed to move an object height h from the earth's surface. so large that:

dent's total work (W) spent a
to move a body with mass m from the earth to a point at distance R from the center of the earth:
$$W=c\cdot m\cdot M\cdot (\frac{1}{R_0}-\frac{1}{R})$$
c = 6.66 * 10 ^ -11, R_0 = 6370
when R increases approaching the term 1 / R all zero, and work to keep a body from the earth's surface infinitely far into the universe can be calculated by the formula
$$W=c\cdot \frac{m\cdot M}{R_0}$$
what I can not really understand is that work is defined as force*distance, W=F*s.
why is $$\frac{1}{R_0}-\frac{1}{R}=distance$$ and why is $$c\cdot m\cdot M=force$$??

can someone explain to me, thanks

2) why is $$c\cdot m\cdot M$$ the same at $$m\cdot g\cdot R^2_0$$, also
$$c\cdot m\cdot M=m\cdot g\cdot R^2_0$$

Last edited: Jan 31, 2010
2. Jan 31, 2010

tiny-tim

Hi aloshi!

(It's not c, it's G. )

Work isn't force*distance unless the force is constant.

Work is the integral of force wrt distance … W = ∫ F.ds,

and in this case F = GMm/r2, so W = ∫ GMm/r2 dr = GMm/r + constant.

3. Jan 31, 2010

aloshi

Re: gravitational

but I can not about Integration, can you explain in a different way? pleas

4. Jan 31, 2010

aloshi

Re: gravitational

unless the force is not constant, way they write the worke sow??

and i find this:
$$W=\int_{r=R_0}^{R}Fdr=\int_{r=R_0}^{R}\frac{cmM}{r^2}dr=cmM\int_{r=R_0}^{R}\frac{1}{r^2}dr=cmM$-\frac{1}{r}$_{R_0}^R=cmM$$\frac{1}{R_0}-\frac{1}{R}$$$$
but i can not anderstund't

Last edited: Jan 31, 2010