-1
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

marked as duplicate by Abhimanyu Pallavi Sudhir, John Rennie, David Z Nov 4 '13 at 9:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/19388 $\endgroup$ – David H Nov 4 '13 at 6:18
  • 2
    $\begingroup$ Comment to the question (v2): In other words, the Moon plays no role in this question other than to fix an initial distance. $\endgroup$ – Qmechanic Nov 4 '13 at 6:55
1
$\begingroup$

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:

Orbit

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.

$\endgroup$
  • $\begingroup$ Time of fall is actually one half of orbital period, so the answer would be about 4.8 days. $\endgroup$ – user23660 Nov 4 '13 at 8:53
  • $\begingroup$ @user23660: oops :-) Thanks - corrected now. $\endgroup$ – John Rennie Nov 4 '13 at 8:58
  • $\begingroup$ Does this take into account acceleration from gravity? $\endgroup$ – Althaen Nov 5 '13 at 5:35
  • $\begingroup$ @Althaen: yes, it's the Earth's gravity that determines the orbit. That's why the Earth's mass, $M$, and Newton's gravitational constant, $G$, both appear in the equation for the time. $\endgroup$ – John Rennie Nov 5 '13 at 7:05
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.