Falling under gravity with decreasing r

[numberjuggler]

If you drop an object from a great height with no air resistance. e.g. 1000 km above the moon. How long will it take to hit the surface?

(GM$_{}1$M$_{}2$)/r$^{}2$ = M$_{}2$a

r$^{}..$=k/r$^{}2$

Hmm this is my first post, not quite getting the symbols.

"The second derivative of r is equal to a constant divided by r squared"

So presumably you integrate that nonlinear differential equation. Somehow.

This is quite an obvious question. It seems like something Isaac Newton would have asked when he first discovered the equations of gravity. A really classic piece of physics. But I can't seem to calculate the answer or find it anywhere.

 

[mfb]

$$\ddot{r}=\frac{-k}{r^2}$$
As code: $$\ddot{r}=\frac{-k}{r^2}$$

Or, in ASCII: d^2/dt^2 r = -k/r^2

The minus sign gives the usual convention that r is the (positive) radius and k is positive, too.

With energy conservation, you can calculate the velocity for every point of your motion. The inverse velocity is "time per length" - if you integrate that over the length of the motion, you get the total time.

 

[MisterX]

Here is a rule which can be useful sometimes.
$\ddot{r} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} =\frac{dv}{dr} v$

Applying this to this situation:
$\frac{dv}{dr} v = - \frac{k}{r^2}$

$v \mathop{dv}= - \frac{k}{r^2}\mathop{dr}$

$\frac{v^2}{2} = \frac{k}{r} + C_1$
We want that at $r_0$, the velocity is zero
$\frac{v^2}{2} = \frac{k}{r} - \frac{k}{r_0}$

Next I used the fact that the velocity of the falling object will be negative.

$v = \frac{dr}{dt} = - \sqrt{\frac{k}{r} - \frac{k}{r_0}}\;\;\;\; r < r_0$

$-\int_{r_0}^{r_1} \frac{dr}{\sqrt{\frac{k}{r} - \frac{k}{r_0}}} = \int_0^t dt'\;\;\;\; r_1 < r_0$
So the time would be
$t = -\frac{1}{\sqrt{k}}\int_{r_0}^{r_1} \frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_0}}} \;\;\;\; r_1 < r_0$
Now comes the task of finding this integral.

$x = \frac{1}{r} \;\;\;\; dx = -\frac{dr}{r^2}$

$dr = -r^2 dx = -\frac{dx}{x^2}$

$-\int \frac{dr}{\sqrt{\frac{1}{r} - \frac{1}{r_0}}} = \int \frac{dx}{x^2\sqrt{x - \frac{1}{r_0}}}$

Many steps omitted. Integral obtained using Wolfram Alpha.

$\int \frac{dx}{x^2\sqrt{x - \frac{1}{r_0}}}$$= \frac{r_0 \sqrt{x - \frac{1}{r_0}} }{ x} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{x - \frac{1}{r_0}}\right)$

substituting back for x:
$= rr_0\sqrt{\frac{1}{r} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r} - \frac{1}{r_0}}\right)$

note that this will in fact be zero at $r = r_0$ so we have

$t = \frac{1}{\sqrt{k}} \left[ r_0r_1\sqrt{\frac{1}{r_1} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r_1} - \frac{1}{r_0}}\right) \right] \;\;\;\; r_1 < r_0$

I am not sure that I have not made a mistake. One thing you could do to check this is to compare it to the falling time for constant acceleration and see if it is close for small distances.

$t = \sqrt{\frac{2(r_0 - r_1)}{g}} = \sqrt{\frac{2r_0^2(r_0 - r_1)}{k}} \;\;\;\; r_1 < r_0$

 

[MisterX]

Sorry, I was missing a $\sqrt{2}$.

$t = \frac{1}{\sqrt{2k}} \left[ r_0r_1\sqrt{\frac{1}{r_1} - \frac{1}{r_0}} +r_0^{3/2}tan^{-1}\left(\sqrt{r_0}\sqrt{\frac{1}{r_1} - \frac{1}{r_0}}\right) \right] \;\;\;\; r_1 < r_0$

 

[PJC01]

Maybe I'm not thinking about it the right way.

I can provide a response to this content by using the principles of physics and mathematical equations. When an object is dropped from a great height with no air resistance, it will experience a constant acceleration due to the force of gravity. This acceleration can be calculated using the equation F=ma, where F is the force of gravity, m is the mass of the object, and a is the acceleration.

In this case, the force of gravity can be expressed as GMm/r^2, where G is the gravitational constant, M is the mass of the moon, and r is the distance between the object and the center of the moon.

By substituting this into the equation F=ma, we can rewrite it as GMm/r^2=ma. Since we are assuming no air resistance, we can also assume that the mass of the object remains constant. This allows us to rearrange the equation to get a=GM/r^2.

Now, we can use the equation for acceleration to calculate the time it takes for the object to hit the surface of the moon. This can be done by using the equation v=at, where v is the final velocity of the object and t is the time it takes to reach that velocity.

Since the object starts with an initial velocity of 0, we can rewrite the equation as v=at=GMt/r^2. We can also use the equation for displacement, s=ut+1/2at^2, where u is the initial velocity and s is the distance traveled, to get s=1/2GMt^2/r^2.

Finally, we know that at the surface of the moon, the distance traveled is equal to the radius of the moon, so we can set s equal to the radius and solve for t. This gives us t=√(2r^3/GM), where r is the radius of the moon.

Using this equation, we can calculate the time it takes for an object to fall from 1000 km above the moon's surface. Plugging in the values for the moon's mass and radius, we get t=√(2(1000km)^3/(6.67x10^-11 Nm^2/kg^2)(7.35x10^22 kg))= 1.4 hours.

Therefore, it would take approximately 1.4 hours for an