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What is the work done by the force of gravity on a particle of mass $m$ as it moves radially from $7500 ~\text{km}$ to $9400 ~\text{km}$ from the center of the earth?

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    $\begingroup$ What work do you have so far? $\endgroup$ – EuYu Mar 28 '13 at 2:08
  • $\begingroup$ do we use $\vec{F} =−\frac{GMm\vec{r}}{\left|\,\vec{r}\,\right|^3}$? $\endgroup$ – Michael Rametta Mar 28 '13 at 2:18
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By the conservation of energy, the amount of work done by working against gravity is the difference in potential energy. That is, the difference in $$ -\dfrac{GMm}{r}\tag{1} $$ between the two points. If we know that $$ \frac{GM}{r^2}=9.80665 \text{ m/sec}^2\quad\text{at}\quad r=6.371\times10^6\text{ m}\tag{2} $$ we get that $GM=3.980\times10^{14}\text{ m}^3\text{/sec}^2$. Now just plug $m$, $r$, and $GM$ into $(1)$.

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  • $\begingroup$ i got 6.224*10^8 i dont think its correct. $\endgroup$ – Michael Rametta Mar 28 '13 at 3:45
  • $\begingroup$ @MichaelRametta: do you know $m$? $\endgroup$ – robjohn Mar 28 '13 at 9:51
  • $\begingroup$ i dont know m thats why i am confused i already know the equation to use i just dont know how to solve it. $\endgroup$ – Michael Rametta Mar 28 '13 at 16:40
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    $\begingroup$ @MichaelRametta: How did you get an answer without knowing $m$? Without knowing $m$, what I get is $$ 3.980\times10^{14}\text{ m}^3\text{/sec}^2\left(\frac1{7.5\times10^{6}\text{ m}}-\frac1{9.4\times10^{6}\text{ m}}\right)=1.073\times10^7\text{ m}^2\text{/sec}^2 $$ which, when multiplied by $m$ in $\text{kg}$, will give the energy in $\text{J}$ (joules). $\endgroup$ – robjohn Mar 28 '13 at 17:17
  • $\begingroup$ from there i would do? $\endgroup$ – Michael Rametta Mar 28 '13 at 22:22
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Hint: $$W = \int F \cdot dx = \int -\frac{GMm}{r^2} \cdot dr = -GMm\int\frac{1}{r^2} \ dr$$

where $M$ denotes the mass of the Earth and $m$ denotes the mass of the object. Integrate and plug in your bounds for the position.

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  • $\begingroup$ what do i do with the given numbers $\endgroup$ – Michael Rametta Mar 28 '13 at 2:21
  • $\begingroup$ Not sure what you mean. Your lower bound and upper bounds on the integral are $7500$ km and $9400$ km. You should know $G, M$ and $m$. Watch your units--I'd convert everything to meters first! $\endgroup$ – Joe Mar 28 '13 at 2:22
  • $\begingroup$ is it 6.224*10^8 ? $\endgroup$ – Michael Rametta Mar 28 '13 at 3:45

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