- #1

- 34

- 0

So there is a hole down through the centre of the earth, spherical earth nothing complicated like air resistance etc.

So basic stuff...

[tex]a = \frac{GM}{r^2}[/tex]

Acceleration of the body due to gravity, where M is the mass of the earth.

SHM

[tex]a=-\omega ^2 r[/tex]

because its displacement is equal to the radius

and the period of a body undergoing SHM

[tex]T= \frac{ 2 \pi }{\omega}[/tex]

So then

[tex]T^2 = \frac {4\pi^2}{\omega ^2} [/tex]

and

[tex]\omega ^2 = \frac {GM}{r^3}[/tex]

Now I'm confused as to why I got the same result as keplers 3rd law

[tex] T^2= \frac {4\pi^2 r^3}{GM}[/tex]

although I probably did something terribly wrong