# Object in orbit vs. object falling through a planet

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1. Jan 30, 2016

### Biker

If you havent seen this video before then go watch it :D
Question:

It is pretty hard to imagine it as a spring and here is why:
1) At (o) the spring should move by its inertia not by any force. If we look at the planet we can summarize that there are 2 points (As I think) that could have this applied to it. First, At the center of the planet, Everything is balanced out and nothing is affecting on it ( Note: I think he mentioned that the gravity is acting from the closest point to r = 0 So the gravity should be at its maximum?). The second point, is really far off the planet (There is still a little bit of gravity)

2) At any of (A) or (B), The force that is acting on the spring should be as big as possible. Well here isnt.. Lets assume that A and B are on the planet's surface( and you have to assume that if you want to calculate the time that the particle will take to cross the planet to the other side) Well as we can see from newton law that:

F = G m M/ r^2

Whenever we go away from an object the force becomes less. Also when the particles goes deeper in the planet the force will become bigger and bigger because you are getting close to the center of (Gravity) force.
And that is not simple harmonic motion.. (From what I studied, Maybe I haven't encountered a lot of states of simple harmonic so I just want to understand this problem)..

So is there is any other explanation or something I missed out in the video?

Last edited: Jan 30, 2016
2. Jan 30, 2016

Staff Emeritus
Maybe you should spend some more time writing your question so that people who might answer don't have to watch twelve minutes of video just to answer you.

3. Jan 30, 2016

### Biker

If you know how to fast forward you will only need 4 mins. Also if write it, It will be a lot boring to read than listen and watch...

4. Jan 30, 2016

### sophiecentaur

This is not right. The only force that's attracting the particle towards the centre of the planet is due to the mass 'beneath its feet'. There is a net attraction of zero from the total mass of the planet that is outside the particle's level. Altogether, that produces a force towards the centre that's proportional to the distance from the centre- =SHM.

5. Jan 30, 2016

### Biker

Oh oh, yeah.. But one thing how there is a net attraction of zero?

What you mean is whenever I go down, I pass some mass(it becomes above me). So it no longer affects on me but what does is what is beneath me.

Last edited: Jan 30, 2016
6. Jan 30, 2016

### PeroK

There are two key results that can be proved for gravity:

1) The gravitational force outside a sphere is equivalent to the force if all its mass were at the centre. That's what allows you to use $F = \frac{GMm}{r^2}$ for a planet, where $M$ is the total mass of the planet and $r$ is the distance from its centre. This is true when $r \ge R$, where $R$ is the radius of the planet.

2) The gravitational force inside a thin hollow sphere is zero.

If you put these two together you find that the gravitational force inside a planet is due only to the mass closer to the centre than you are.

You might want to play about with these equations for a planet of uniform density and see whether you can derive the equation for SHM.

7. Jan 30, 2016

### sophiecentaur

The Shell theorem tells us just that.
There's an arm waving argument to show that it 'could' be true. Sit on a point on a diameter of a spherical shell. Looking at the closer and further surfaces and imagine a narrow cone, projecting discs on both internal surfaces. Their areas are proportional to the distance from it squared and the ISL tells you the force from a disc is inversely proportional to the disance squared. Look in one direction and you see a small disc, close to you and, in the other direction and you see a larger disc that's further away. The area will be bigger but the distance will be greates (squared and 1/squared) so the resulting forces will balance.

If the planet is uniform density, the oscillation period would be the same for all starting heights (and for all diameters of planet. The period of oscillation for the Earth is about 90 minutes. You have to assume the Earth has uniform density (which it doesn't - but never mind). Also ignore the vast pressures and temperatures inside and the spin problem. You could take an asteroid of the same material (or even a small spherical rock) with a hole through it and drop a pebble through a hole through it and you would get the same 90 minute oscillation period.

8. Jan 30, 2016

### Biker

I guess that is a bit much for me...
The best I could imagine that the force becomes lower and lower because the mass behind you pulls so it the decreases the net of the force...