Question

If a stationary object of 1kg starts 100000m from the earth (mass ~6*10^24 kg), what time will it hit the Earth's surface (~6000m from the centre of mass)? More generally, though- what is the equation linking the distance and time in Newton's law of universal gravitation?

Answer 1

Consider the path of the object in terms of one spacial variable: radial distance. Then
\[ F = ma = m \frac{d^2r}{dt^2} = -\frac{GMm}{r^2} \] \[\implies \frac{d^2r}{dt^2} = - \frac{GM}{r^2} \]

Together with initial conditions
\[ r'(0) = 0, r(0) = 100,000 \] we have a non-linear, second-order initial value problem, the solution to which will give you the time you're asking for.

The annoying thing about this problem is I don't think there is an analytic solution. I'd be very happy is someone thinks otherwise, as this is a problem that I'd like a neat answer to as well.

Answer 2

Thanks, that was much as I feared. Odd how calculus breaks sometimes.

Answer 3

r''=G M /r^2

can't we just integrate it twice?

Answer 4

Ok, so I've spent some time I'm playing around with this and it can be done. I'm surprised this didn't turn up as a first-year Physics question. Maybe I goofed off that week. Anyway here goes:

Starting with
\[ r'' = - GM/r^2 \]
multiply both sides by r' dt
\[ r' r'' dt = -GM/r^2 r' dt \]
Now, r' dt = dr and r'' dt = d(r'), hence we have
\[ r' d(r') = -GM/r^2 dr \]
and integrating
\[ \frac{1}{2} ( r'^2(t) - r'^2(0)) = GM \left( \frac{1}{r(t)} - \frac{1}{r(0)} \right) \]
Write \( r_0 = r(0) \) (this is the 100,000 m in the original problem), \( r = r(t) \) and notice that \( r'(0) = 0 \); thus
\[ r'^2 = 2GM \frac{ r_0 - r }{r_0 r} .\]
Now ...

Answer 5

...taking the negative square root as r is decreasing, we have
\[ \sqrt{\frac{r_0}{2GM}} \sqrt{\frac{r}{r_0 - r}} \ dr = - dt \]
\[ \sqrt{\frac{r_0}{2GM}} \left( - r_0 \arccos \sqrt{r/r_0} - \sqrt{r(r_0-r)} \right) = - ( t(r) - t(r_0) ) \]
(*This integral on the left required a bit of thought, deciding between arccos and arcsin; we'll see in a moment that the choice of arccos makes physical sense.)

On the right, \( t(r_0) \) is the start time and is equal to zero. Therefore

\[ t(r) = \sqrt{\frac{r_0}{2GM}} \left( r_0 \arccos \sqrt{r/r_0} + \sqrt{r(r_0-r)} \right) \]
Notice when \( t(r=r_0) = 0 \). We have also written this 'stopping time' \( t \) a function of \( r \), the end point radial distance. But we can also consider it a function of the starting radial distance, \( r_0 \). Thinking of the function that way, notice now that \[ t(r_0) \rightarrow \infty \ \hbox{ as } r_0 \rightarrow \infty. \]
This also makes physical sense as the initial acceleration of a body a long distance from a single body will be very low.

Answer 6

Writing finally \( r = R \) the radius of the earth, the answer to the initial question is
\[ t(R) = \sqrt{ \frac{r_0}{2GM} } \left( r_0 \arccos \sqrt{R/r_0} + \sqrt{R(r_0-R)} \right) \]

Answer 7

So let's calculate: let
\[ r_0 = R + h \]
If an object starts 100 km above the earth, h = 100,000 m and
\[ r_0 = 6,471,000 m \]
(note, this means what I wrote above about r0 isn't right; we need to add the radius of the earth)

For this value, the time to reach the earth is 144 sec.

Running the calculation for difference distances:
h = radial distance from earth's surface (now measured in km)
t = time to reach surface (seconds)

h t
1 km 14.3 sec
10 48.2
100 144
1000 510

At the distance of the radius of the moon's orbit, h = 385,000 km, the time is
\[ t = 430,300 \ sec = 119.5 \ hours \]

Attached is a plot of time (y axis, seconds) as a function of starting distance, h (x axis, kms)

Answer 8

What's very interesting about the time function is there is an inflection point and for very large distances, the time required to reach the surface increases at an increasing rate; i.e., positive second derivative.

Here's the same graph again, now going out to 1,000,000 kms from the earth's surface. Vertical axis is still seconds. Note the shape of the curve.

If you'd like to play with this yourself, here's the formula in Wolfram:
http://tinyurl.com/7gu8l7b

Answer 9

Very nice, I'll play with that later. Thank you very much! I was also trying to figure this one out but I wasn't able to do it properly.

Answer 10

Or you can just use kepler's laws of planetary orbit T^2 proportional to R^3 just you need to show that the radial infall coresponds to a degeerate ellipse

Answer 11

I'd be impressed if that works. Show us how Kepler's law reproduces the formula above for time.

Answer 12

If the earth is taken to be a point object which it can be for the case use kepler's law of periods and put r=100000/2 and calculate time of period to go around and return to same point since it doesn't return time taken to fall must be half of total by symmetry...
You can escape calculus if you are lucky enough to find a point object

Answer 13

James J, I have a couple of questions (due to my lack of a good calculus knowledge) with regards to your solution:
-Is it possible to multiply both sides by dr? I was always told that the maths didn't let you multiply both sides by an infinitesimal.
-In the 4th step, what did you integrate the 2 sides with respect to? It seems as if it was \[d(r ^{\prime})\] on the left side and dr on the right.

Answer 14

- You can, if you think of them as something called 1-forms. This is something my HS mathematics teacher would have deduced marks for. But later on in University, you can do this if you understand exactly where this objects sit. In fact, even if you don't, you get familiar with manipulating these sorts of expressions in the first ODE course you do.

- Yes, exactly. Both sides had different variables of integration. One was r (notated by dr) and the other was r' (notated by d(r')).

Answer 15

OK, Thanks very much, that was a great help.