- #1
Blistering Peanut
- 34
- 0
I was just trying to figure out what would its period be and see where I went wrong (not homework, more of a thought experiment).
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
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