### Related Questions

out of the 15 stats of matter how many can be found on earth and where can the others be found

Can black holes be used to prevent the heat death of the universe?

what planets closer enviromentally to earths (doesnt have to be in this solar system )i know of one that is similar but

how far is earth from the sun?

In a black hole, do electrons overlap each other?

## Comments

freddiecommented on :Thinking about this initially, I thought simple harmonic motion. But then I thought more about it.

In SHM, the acceleration is maximum at the amplitude, however, Newton’s law of Universal Gravitation states that gravitational force (hence acceleration) decreases as distance from the centre increases. Therefore I have a contradiction – it can’t be SHM.

The equations of motion I formed are as follows:

-GM/x^2 = x” if x is positive

GM/x^2 = x” if x is negative

x=0, x”=0

Where x is the distance from the centre of the earth in the starting direction. M is the mass of the earth.

Due to the symmetry of the equations, the ball would go back and forth like SHM but the motion wouldn’t satisfy the SHM equation.

The graph would look something like this: http://i.imgur.com/gApTJ.png . I could only get Autograph to do the positive values (it got confused when x=0!). (For simplicity, GM = 1 and earth radius = 3, x is on y-axis, time on x-axis).

As the ball approaches the centre of the earth, its acceleration approaches infinity. However, luckily once it reaches the centre, the acceleration has no direction so cannot exist.

dopplercommented on :At first I also thought the acceleration would increase as the ball approaches the centre, but then (this made me think hard!) I realised this wouldn’t be the case. We can see this if we model the Earth as a sphere of radius (and so distance from the centre) x, and mass M, with an equal distribution of mass throughout. This means its mass, M, is proportional to the volume of the sphere, so M is proportional to (4πx^3)/3, so (as 4π/3 is a constant) M is proportional to x^3. In fact, the part of the earth at a distance from the centre greater than where the ball is at any given time has a net force of zero on the ball, because it forms a hollow sphere and any object within a uniform hollow sphere experiences no net force from it, no matter where the object is inside it. This means we can ignore any part of the Earth that is further from the centre than the ball itself is, and only consider the force the new, smaller and less heavy “Earth” (whose surface is at the same distance from the centre as the ball is) exerts on the ball. According to Newton’s Law of Gravitation, F=GMm/(x^2); Gm is a constant as we are not changing the mass of the ball, so F is proportional to M/(x^2). M is proportional to x^3, as shown above, so F must be proportional to (x^3)/(x^2). This means F is directly proportional to x. As F=ma, acceleration is also proportional to x, so this really is an instance of s.h.m. (as s.h.m. is motion where acceleration towards a point is directly proportional to the distance from that point).

Hope that that was helpful!

😀

freddiecommented on :But the mass M is constant and doesn’t change as the ball moves towards the centre. So I still don’t think it’s SHM.

dopplercommented on :The mass of the sphere with a radius the same as the distance the ball is from the centre at any stage is the only thing which contributes a value to the net force on the ball – all the matter further away has a net force of zero on the ball (if you’re finding it hard to visualise this, don’t forget there is some matter above the ball which is pulling it backwards!) See Sheila’s awesome answer above: there is actually no net force on the ball at the centre, hence no force on the ball. Thinking about it another way, an atom which is a nanometre away from the centre of the Earth certainly isn’t being accelerated towards the centre by a practically infinite force… which the model you put forward would calculate.

dopplercommented on :I’ve heard that Newton was very interested in this problem, and that the ball’s path would depend on whether it was dropped at the pole or at the equator due to the motion of the Earth spinning around its axis – is this true?

Suziecommented on :Good point, i hadn’t thought about the effects of where you drop the ball, I’d only considered a static Earth, that would complicate things quite a bit as well!

You guys have done so well asking questions, if you have twitter I’m @suziesheehy if you want to keep asking/discussing after the competition.

ddkcdkcommented on :Im liking all of your answers here. I went along the same sort of lines when i was asked it…

I started talkin about shm and worked out a few numbers in my head, (tried to look clever;) lol). Then talked about the different scenarios like if there were any resistive forces etc. In the end we just left it there because there was no right or wrong answer to it (and he didnt wanna spend all the interview on one question!)