Drilling through the Earth problem

[dinospamoni]

Homework Statement

Postal workers on planet Vashtup want to drill a
straight tube through the planet, starting at Post Office 1, passing through the center of the planet, and ending on the other
side at Post Office 2. They plan to release small packages containing mail into the tube from P.O. 1 and have others grab
them at P.O. 2. Vashtup has g = 9.1 m/s2
, and a radius of
5200 km. When it is located within the shell of a planet, the
weight of a particle of mass m is mgr/R, where r is its distance
from the center of the planet. Assume that there is no air resistance. Compute a) the position r of the package 1088 s after
it has been released, and b) its speed at that time. Note: r is
positive if the package is on the same side of planet as P.O. 1,
and negative if it is on the same side as P.O. 2.

Homework Equations

F=ma

The Attempt at a Solution

Note: My professor wants me to be using differential equations to solve this.

I think I have a basic idea of how to do this problem, but am having trouble finding an equation that correctly models the motion of the object dropped into the tube. From what is given I have:

ma=mgr/R

a=gr/R

a is the second derivative of position, so x''=gr/R

Now, this is where I'm having trouble. I know I need to find r(t), or a function for position dependent on time but I'm not seeing how to make that connection.

I tried using the above equation saying (R/g) r''(t) = r(t), y'(0)=0, y(0)=R

but that definitely isn't right

Any ideas?

 

[rude man]

Homework Statement

I tried using the above equation saying (R/g) r''(t) = r(t), y'(0)=0, y(0)=R

but that definitely isn't right

Any ideas?

Well, I don't know what those y's are doing there, might want to change those to r's, but if you rethink the signs of your equation I would not say it's 'definitely not right' because it is!

Hint: You're sitting at r = R to begin with so think about F = ma with the correct sign (direction) in your coordinate system.