Ignoring the moons gravity, if an object sitting still (relative to the Earth, i.e. not in orbit) was dropped from the moon. How long would it take to hit the Earth?
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The equation of motion for an object falling radially towards the Earth is a bit involved to integrate (see the duplicate question links above for details) but there's a sneaky trick to get the time. Look at this diagram:
This shows the orbit of the Moon round the Earth (approximated as a circle) and the orbit of the dropped object assuming some small lateral velocity. The object traces a very long shallow ellipse, and the period of an elliptical orbit is:
$$ T = 2\pi \sqrt{\frac{a^3}{MG}} $$
where $a$ is the semi-major axis and $M$ is the mass of the central body. Note that the eccentricity of the ellipse doesn't change the period, so this equation gives the correct period even in the limit of the ellipse becoming a straight line.
So just plug in $a = 1.922 \times 10^8$ (half the Earth-Moon distance) and take half the period and we get:
$$ T \approx 419,000 \text{seconds} $$
or about 4.85 days.
John Rennie
- 367,598
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It would quite likely never reach Earth. This is because the Moon is in orbit around the Earth, so if you took away the Moon's gravity then any object on its surface would have sufficient velocity to be in orbit around the Earth in its own right.
N. Virgo
- 35,274
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Given the regular notations for gravitational force, $$ F = G \frac{Mm}{r^2} $$
The acceleration of the object would be $$ a = \frac{GM}{r^2} $$
Consider that for an infinitesimal distance $dr$, the acceleration is constant. This would give
$$ v_0 = u_0 + a~dt\\ \therefore dt = \frac{v_0 - u_0}{a} = \frac{(v_0 - u_0)r^2}{GM} $$
where,
$$ v_0 = \sqrt{u_0^2 + 2 \frac{GM}{r^2} dr } $$
Integrating the resulting quadratic differential equation over r = (distance between centers of the moon and earth) to the radius of the earth, you have your answer.
pho
- 4,665