Question: If you dig a hole right through the earth, and drop a ball through it, what will happen to it? ( I got asked this question in my oxford interview practice...lol. So just wanna see what you think?)
If there was no air resistance (or any other effects from variation of the density of the Earth’s core), the ball would leave your hands, accelerating towards the centre of the earth – and will keep accelerating until it reachest it’s maximum velocity as it passes the centre of the Earth. Once it passes that point, it would start decellerating – and would reach a stop at the other side of the earth – it would then accelerate back towards the centre of the Earth, and so on and so forth – simple harmonic motion! Of course there are other factors in reality which would mean this wouldn’t be viable – like drilling a hole through the Earth!
I think the answer is this (though maybe look it up if you think you’ll really get asked that as an interview question, I wouldn’t want to feel responsible!!):
The ball will be travelling through air so at some point when you drop it will reach terminal velocity. As it reaches the centre the gravitational force on it is smaller and so after it passes the centre it will oscillate about the centre and eventually come to rest at the centre of gravity. (If there were no air I would expect it would go through & out the other side to the opposite of where you dropped it and then oscillate the full way back and forth through the Earth without coming to rest.)
The easier answer is that it will melt as it approaches the centre and get stuck to the sides of the hole you’ve drilled :-p
The ball will initially accelerate at 9.81 m/s^2 towards the center of the Earth due to gravity. But as it picks up velocity, its acceleration will decrease because more and more mass will be above the ball and less and less below it. The increasing mass will cause an opposite effect to gravity and that will offset the decreasing gravity pull down.
The net gravity force will go to zero as the ball approaches the center of the Earth.
Now we have to imagine that the Earth is cool in the middle because otherwise the ball will just vaporise with the temperature and the pressure. When the ball reaches the center of our imaginary cool Earth, it will have reached about 28000 km/h, and the mass above it would equal the mass below it. This means the force of gravity will be pulling it equally from all around, so the gravity forces cancel out and there is no net force acting on the ball when it’s dead center. Acceleration at this point will be exactly zero.
But the ball’s momentum p = mV will be at its maximum. With no forces acting on it to stop the ball, it carries on right through the center and starts its rise to the other side of the Earth. As it ascends, it’s acceleration, which is now a deceleration, increases at the same rate it decreased when heading toward the center.
In other words, the ball will follow a mirror image of the trajectory it travelled when heading inbound to the center. Instead of increasing velocity, the initial 28000 km/h velocity decreases at increasing deceleration as more and more of Earth’s mass is behind it and less and less is above it. Finally, when it reaches the surface on the other side, it will have reached zero velocity with an acceleration of 9.81 m/s^2 back toward the center of the Earth, but from the other side this time.
And if no one is there to catch that ball as it pops up on the other side, it will fall back down into the hole and start the journey all over again!!
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
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.
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).
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.
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?
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.
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!)