Calculation of when Moon will hit Earth if not orbiting?

Question

Me = 5.97 x 10^24kg
Mm = 7.34 x 10^22kg
r=3.84x10^9m

F = G*Me*Mm/r^2
F = 6.67x10^-11(5.97 x 10^24)(7.34 x 10^22kg)/(3.84x10^9)^2
F = 1.98x10^18N

F = Mm*a
1.98x10^18N = 7.34 x 10^22(a)
a = 2.7x10^5m/s^2

v2^2 = v1^2 - 2ad
v2^2 = 0 - 2(-2.7x10^5)(3.6x10^9)
v2^2 = 1.94x10^15
v2 = 4.41x10^7m/s

a = v2 - v1 / t
2.7x10^5 = 4.41x10^7 / t
t = 163sec

SO, if Earth and Moon were completely still and if the moon wasn't orbiting the Earth, the Moon would hit Earth in 163 sec which is just under 3 min. But i don't know if I was correct in my calculations to make Earth a point reference.

Now the question, is the Moon's orbit around the Earth (constant free-fall) preventing the Moon from hitting Earth??

Answer 1

Your calculation is TOTALLY SCREWED UP --- a technical term that we in the astrophysics trade use.

1. Your calculation of the instantaneous value of the INITIAL acceleration 'a' was TOO LARGE by 10^10 --- your EXPONENT of 10 should have been - 5, not + 5.

(You should have realized that an answer of ~ 163 seconds for the Moon to fall towards the Earth with its initial acceleration was ABSURD.)

2. And, of course, the Moon would experience an INVERSE SQUARE LAW of attraction, so that the attraction would speed up as time progressed and it got nearer to the Earth.

What one can show with a proper approach is that it should take:

(sqrt 2) / 8 months to hit the Earth,

starting from rest at its mean distance, that is:

0.176777... months = 4.83... days to hit the Earth from rest.

And the answer to your final question is of course: YES!

Live long and prosper.

P.S. Note that only physicists would integrate --- twice --- to obtain the answer, and thus make rather heavy weather of it. For an astronomer, the problem can be solved in one or two lines of simple algebra, by appealing to Kepler's 3rd law applied to a suitable --- if degenerate --- orbit. That's how the answer of 1 / [2 sqrt 2] or (sqrt 2) / 8 of a month for the time of fall was obtained. (The corresponding time for the Earth to hit the Sun from rest is (sqrt 2) / 8 of a year, that is ~ 64.5 days or slightly more than two months.

Also, as Greek astronomers knew, the Moon's orbit is about 60 times the radius of the Earth, not just 36. Thus Newton's celebrated "Moon test" was in fact a check on whether the gravitational acceleration at the Moon's distance was ~ 1/3600 of its value at the surface of the Earth. Note one rather amusing coincidence because of that factor of 60 in the distance of the Moon (or 1/3600 in the acceleration there), and the fact that there are 60 seconds in one minute. That is, that the Moon has an acceleration of ~ 32 ft or ~ 9.8m per (minute)^2 towards the Earth rather than the near-Earth apple's ~ 32 ft or ~ 9.8m per (second)^2 !!

The practical consequence of this latter fact is that the Moon "falls" just as far towards the Earth in one minute as an apple here falls in one second! Amazing coincidence of how the numbers work out, but true.

Answer 2

Wrong. The moon currently moves away from Earth, by 4 cm per year, because of tidal locking. It will stop the movement away from Earth when the orbit period of the moon and the rotation of Earth (=length of our day) are equal to 49 days... in about 5 billion years. The moon was about 4 billion years ago a part of our planet according to the best theory about it's creation, until a planetoid (or protoplanet) of the size of mars collided with Earth. The solar system consisted of about 100-200 dwarf planets at that time, with such collisions being common. The collision with the object, called Theia today, destroyed Earth and Theia almost completely and debris of the collision formed the moon - initially in pretty low orbit. Earths gravity distorted this object and tidal forces between both slowed Earths rotation down from 6 hours to 24 hours today and send to moon away to it's current position.

Answer 3

As a sanity check, calculate the time it would take an object to fall the distance at 1G. But first, note that your distance is too large by a factor of 10. That time comes out to about 2.5 hours. But that's the final acceleration. The initial acceleration is much less, since the starting distance is around 36 times the ending distance, so the initial acceleration is 1/1360 of the final acceleration. You would need to take an integral to get the answer, but it will be a lot longer than that 2.5 hours.

Answer 4

Let's do it as a 1-dimensional problem: straight drop of the Moon onto the Earth:
- mass of Moon: m
- mass of Earth: M = 5.97e24 kg
- distance between centers: r
- initial distance: R0 = 3.84e8 m [I think you got this off]
- final radius of Earth-Moon separation: sum of the two radii:
R1 = 1.74e6 + 6.38e6 = 8.12e6 (m)
- gravitational constant: G = 6.67e-11

OK, let's go:
-GMm/r^2 = mr'' ; multiplying by r' on both sides, and canceling m:
-GMr'/r^2 = r''*r' ; noting that both can be integrated wrt time:
+GM/r = (r')^2/2 + constant.
Since r' = 0 when r = R0,
GM/r - GM/R0 = (r')^2/2 , or
r' = - sqrt(2GM(1/r - 1/R0))
= - sqrt(2GM)*sqrt(1/r - 1/R0)
=>
dr/sqrt(1/r - 1/R0) = - sqrt(2GM)*dt
r^(1/2)dr/sqrt(1 - r/R0) = - sqrt(2GM)*dt

Integral(r=R0,R1)[r^(1/2)dr/sqrt(1-r/R0)] = - sqrt(2GM)*t
If you play around with this long enough, you will eventually stumble across the following substitution: define (cos(y))^2 = r/R0, and the integral becomes:
- R0^(3/2) *2 * Integral(y=0, Arcos(sqrt(R1/R0))) [(cos(y)]^2 dy]
= - R0^(3/2) * Integral(y=0, Arcos(sqrt(R1/R0))) [(1 + cos(2y)] dy]
= - R0^(3/2) * [Arcos(sqrt(R1/R0)) + sin(2*Arcos(sqrt(R1/R0))/2)

Therefore,
T = [R0^(3/2)/sqrt(2GM)] (Arcos(sqrt(R1/R0) + (1/2)sin(2*Arcos(sqrt(R1/R0)))
= [R0^(3/2)/sqrt(2GM)] (Arcos(sqrt(R1/R0) + sqrt(R1/R0) * sqrt(1 – R1/R0))
= [(3.84^(3/2)*e12)/sqrt(2*6.67e-11*5.97e24)]*[Arcos(sqrt(8.12e6/3.84e8))
+ sqrt(8.12e6/3.84e8))*sqrt(1-8.12e6/3.84e8]
= [3.84^(3/2)/sqrt(20*6.67*5.97)]*e6*[Arcos(sqrt(8.12/3.84)e-2)
+ sqrt(8.12/3.84)*(e-1)(1-8.12/384)]
= [2.66e-1]*e6*[Arcos(0.145) + 0.145*(1-0.02115)]
= [0.266]*[1.425 + 0.419]*e6
= 4.9e5 seconds
= 5.67 days
This is a bit different from what Dr Spock gets, 4.83 days, but in the same range. There could easily be an arithmetic error somewhere in there. The fact that he doesn't take into account the finite radius of the Earth & Moon shouldn't make any difference.

Answer 5

is that supposed to be impressive? why don't you calculate what has to happen, in order for the earth and moon to be stationary? come back in about 400 years and let me know.