# Falling body where acceleration is not constant

1. Mar 23, 2015

### barryj

1. The problem statement, all variables and given/known data
Most physics problems that involve a falling body assume constant acceleration. How does one account for the changing acceleration if a body is dropped from a very high altitude?

2. Relevant equations
With constant acceleration, d = (1/2)at^2 or t = sqrt(2d/a)
but a is not really constant so How is this solved.

3. The attempt at a solution
My first thought was to do a piece wise solution. Start at a given altitude, assume a small time, say one second and calculate the distance traveled. Then calculate the acceleration based on the new distance between bodies and continue. Is there a closed form solutio?

2. Mar 23, 2015

### PeroK

Is this really homework?

Anyway, the direct approach involves solving a differential equation.

Alternatively, you could consider conservation of energy.

3. Mar 23, 2015

### barryj

No, this is not homework but rather a problem I would like to know the answer to. I figured that a diff equation would be required but I don't see the trick in solving it. Ultimately, I would like to get the altitude as a function of time but since the acceleration is a function of altitude I am not sure how to set this up.

4. Mar 23, 2015

### PeroK

Suppose an object of mass $m$ starts at rest from a height $h$. How long does it take to fall, using a varying gravitational force?

You can start with energy conservation using gravitational potential. Let $r$ be the distance from the centre of the Earth, $M$ the mass of the earth, and $R$ is the radius of the Earth.

$PE(r) = -\frac{GMm}{r}$

$KE(r) + PE(r) = PE(R+h)$

$\frac{1}{2}m(\frac{dr}{dt})^2 = \frac{GMm}{r} - \frac{GMm}{R+h}$

That gives you a differential equation in $r$.

The trick to solving it is to let $r = (R+h)cos^2(\theta)$

5. Mar 23, 2015

### barryj

I understand how you used the conservation of energy to get the differential equation.
However, I seem to need another hint to see how to use the substitution r = (R+h)(cos A)^2

6. Mar 23, 2015

### PeroK

Just plug it in and simplify. For example:

$\frac{dr}{dt} = -2(R+h)cos(\theta)sin(\theta)\frac{d \theta}{dt}$

As you can see, things get worse before they get better!

7. Mar 23, 2015

### barryj

Looks like you can integrate to get r = (R+h)(sin A)^2 but then using the change of variable for limits gets messy.

Maybe my numerical approach might be better to get a velocity at a specific height but I wold have rather get a closed form solution.

I see why this is not a Physics homework problem.

8. Mar 24, 2015

### PeroK

It's not too bad. We have:

$\frac{1}{2}m(\frac{dr}{dt})^2 = \frac{GMm}{r} - \frac{GMm}{R+h}$

$(\frac{dr}{dt})^2 = 2GM\frac{R+h-r}{r(R+h)}$

Letting $r = (R+h)cos^2(\theta)$ gives

$4(R+h)^2cos^2(\theta)sin^2(\theta)(\frac{d \theta}{dt})^2 = \frac{2GM(R+h)(1 - cos^2(\theta))}{(R+h)^2 cos^2(\theta)}$

Which simplifies to:

$cos^2(\theta)(\frac{d \theta}{dt}) = \sqrt{\frac{GM}{2(R+h)^3}}$

We can then integrate from $t = 0, \ \ r = R + h, \ \ \theta = 0$ to $t = T, \ \ r = R + z, \ \ \theta = \theta_z$

Which gives the time $T$ taken to fall from a height $h$ to a height $z$

$T = \sqrt{\frac{(R+h)^3}{8GM}}(2 \theta_z + sin(2 \theta_z)) \ \ (\theta_z = cos^{-1}(\sqrt{\frac{R+z}{R+h}}))$

To get the time to fall to the ground, you just set $z = 0$

I checked this formula out and found that for a fall of 1,000m the difference is only 0.002s compared to using the constant surface gravity.

9. Mar 24, 2015

### barryj

Wow! I would never have solved this problem. The real reason I wanted to solve his problem is that I was reading a si fi novel where an asteroid caused the moon to stop orbiting around he earth. I wondered how long it would take for the moon to collide with the earth if it were not moving in its orbit. Then I wanted to know what an observer would see as the moon got closer and closer to the earth. Now I can figure it out.

thanks

10. Mar 24, 2015

### PeroK

For something as far as the moon, you can get an estimate by taking $R = 0$. In which case:

$T = \pi \sqrt{\frac{h^3}{8GM}}$

Where $h$ is the distance to the moon, give or take the Earth and moon radii.

11. Mar 24, 2015

### barryj

I calculate 4.76 days. Now for the final question. What would be the relationship between distance or height and time, i.e. distance = f(time)?
From this I could calculate the angular size vs time. It obviously would not be parabolic as the acceleration is continually changing.

I recall a movie recently where some planet was going to hit the earth and the movie ended with the star, Kirsten Duntz as I recall sitting on her and watching patio as the approaching planet got slowly larger and larger.

12. Mar 24, 2015

### PeroK

You can get it from that equation. We can set $R = 0$, then we have:

$T = \sqrt{\frac{h^3}{8GM}}(2 \theta_z + sin(2 \theta_z)) \ \ (\theta_z = cos^{-1}(\sqrt{\frac{z}{h}}))$

That gives you the time at distance $z$.

You could use a spreadsheet to calculate $T$ for whatever set of distances you want.

13. Mar 24, 2015

### barryj

Will do. thanks for your help.