Question

Suppose that a hole is drilled through the center
of Earth to the other side along its axis. A small
object of mass m is dropped from rest into the
hole at the surface of Earth, as shown above.
If Earth is assumed to be a solid sphere of mass
M and radius R and friction is assumed to be
negligible, correct expressions for the kinetic
energy of the mass as it passes Earth’s center
include which of the following ? I know the answer but how do you solve for the answer?

Answer 1

The answer is E.

Answer 2

@IrishBoy123

Answer 3

@IrishBoy123

Answer 4

I know that kinetic energy is \[(1/2)mv^2\]

Answer 5

And that the gravitational potential energy is \[\frac{ GMm }{ r }\]

Answer 6

gravity decreases linearly from g_s at the earth's surface to zero at its centre. it can be compared to a spring, ie as if there is a spring at the centre of the earth pulling the mass m toward it.

\[F = mg_s \frac{r}{R} = \frac{mg_s}{R} r = k \ r\ ; \ k = \frac{mg_s}{R} \]

the potential energy in the "spring" at the earth's surface will be converted into kinetic energy at the earth's centre. said energy is:\[\frac{1}{2} k x^2 = \frac{1}{2} \frac{mg_s}{R} R^2 = \frac{1}{2} \ mg_s R\]

and, if you fiddle about with the idea that
\[F = \frac{GMm}{R^2} = mg_s\]you can replace g_s with:
\[\frac{GM}{R^2} \]
to perhaps find an additional answer....

Answer 7

I understand that the force due to gravity at the surface is F = mg but where did the (r/R) come from?

Answer 8

first to scope this out, imagine you are at a cavity at the exact centre of the earth. gravity is pulling radially outward as all the mass is between (a) you at the earth's centre and (b) the earth's surface. you will realise that due to symmetry, there net gravity should be zero. that means that gravity varies between g at the earth's surface and zero at it's dead centre.

the question is - how does it vary with radius. it does not have to be a linear relationship, though in fact it is.

to show this, we can move the cavity away from the centre of the earth and place it anywhere within the earths surface. and then we split the earth into 2 regions, as per the drawing.
it turns out[see below] that we can, in order to calculate gravity at radius Ri, completely ignore the gravity component of the outer shaded shell - it nets out to zero. this means that gravity at r = Ri, the new position of the cavity, is \[F = \frac{GMm}{R_i^2} = mg_i; \ G = \frac{g_i R^2_i}{M_i} = \frac{g_i R^2_e}{M_e}\]
\[\frac{g_i R^2_i}{\frac{4}{3} \pi R^3_i} = \frac{g_i R^2_e}{\frac{4}{3} \pi R^3_e}; \ \frac{g_i}{R_i} = \frac{g_e}{R_e}\]
thus \[g_i = g_e \frac{R_i}{R_e}; \ ie \ linear\]

obviously, the big idea here is ignoring the outer shell. to show that we can ignore the outer shell requires some calculus, which is not as bad as it looks, but would begin to turn this post into a book. we can do that too if you are interested. i am.

you can also use Gauss' Law, ie the gravity analogue of Gauss' Law for electrostatics. [but that's a "cop out" too (!!) as you are relying on the Gauss "black box".]