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Find the minimal velocity needed for a meteorite of mass $m$ to get to earth from the moon. Hint: the distance between the center of earth and the center of moon is $\approx 60 R_E$, and the meteorite should reach a certain point $O$ on that distance, where the gravitational forces of the moon and the earth are canceling each other out. It is located $\approx 6 R_E$ from the center of the moon.

I thought that the best way to solve this question is to find the difference between the initial enegy of the body and the final one at $O$. Beyond that point, the gravitational pull of the earth will 'overcome' the gravitational pull of the moon, and the object will gather speed itself by reducing the GPE in the earth's gravitational field. Therefore:

$U_{G,i}=-G \frac{M_{moon}m}{R_{moon}}$

$U_{G,f}=-G \frac{M_{moon}m}{6R_E}-G \frac{M_E m}{54R_E}$

$U_{G,f}-U_{G,i} \approx m \cdot 1.51 \cdot 10^6 \text{J}$

(there's no mistake in the calculation)

$E_k=m \frac{v^2}{2}=m \cdot 1.51 \cdot 10^6$

Therefore $v \approx 1739.97 \text{m/s}$

However this answer is wrong. The correct magnitude of the speed should be somewhere around $2.26 \text{km/s}$ which is close to the escape velocity from the moon ($\approx 2.37 \text{km/s}$).

Where am I wrong? Why the difference in GPE is not satisfying the problem?

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$$U_{g,i}=-G\dfrac{M_{moon}.m}{\underbrace{R_{moon}}_{(\ distance \ from \ moon\ center) }}-G\dfrac{M_{Earth}.m}{\underbrace{(60R_{earth}-R_{moon})}_{(initial \ distance\ from\ Earth)} } $$

You missed $U_i$ due to Earth!

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  • $\begingroup$ But the meteorite is initially located on the moon. Do you imply that I should take into account the gravitation of earth when the object is placed somewhere on the moon? $\endgroup$
    – grjj3
    May 2, 2013 at 18:37
  • $\begingroup$ Aah yes! I got it reversed. I thought it to be on earth initially.@grjj3 Now I have Edited it This is final expression. $\endgroup$
    – ABC
    May 2, 2013 at 18:37
  • $\begingroup$ Thank you both very much for your help. And even more - for the patience on my sometimes stupid questions. $\endgroup$
    – grjj3
    May 2, 2013 at 18:44
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The initial kinetic energy needed is whatever it takes to get from the start to the maximum potential between the start and destination.

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  • $\begingroup$ That's what I did, but my answer is too far from the correct one. $\endgroup$
    – grjj3
    May 2, 2013 at 18:22

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