# Math Help - Tricky Physics Proof (that g=9.8)

1. ## Tricky Physics Proof (that g=9.8)

Prove that the gravitational field strength on earth is 9.8N/kg given ONLY the following:

Period of the moon around earth is 27 days, 8 hours.
Earth's radius iis 6.38 x 10^6 m
G = 6.67 x 10^-11 Nm^2/kg^2.

Full solution is needed

If anyone could help me with this, it is a "fun" challenge.

2. First, we set the centripetal acceleration equal to the gravitational force to get: $\omega^2=\frac{GM}{r^3}$. To relate this to a period, we note that $T=\frac{2\pi}{\omega}$, and solving for $M$ we get $M_\oplus=\frac{4\pi^2r^3}{GT^2}$. This gives us an expression for the mass of the earth given the period and radius of the moon’s orbit.

Then, we simply substitute into an expression for $g=\frac{GM_\oplus}{R_\oplus^2}$, where $M_\oplus$ is the mass of the earth, and $R_\oplus$ is the radius of the earth. Thus, $g=\frac{4\pi^2r^3}{T^2R_\oplus^2}$.

So you don’t even need to know $G$!

3. Thanks for your solution, but I think that it is supposed to be done a different way for this challenge. We were definitely supposed to use G. The only equations that can be used are:

Fg = mg , F(G) = G(m1)(m2)/d^2, F(net) = ma, and Fc = 4(pi^2)r /T^2

Unless your equation for centripetal force is somehow derived from ^ that one

4. Originally Posted by dabigchz
Fc = 4(pi^2)r /T^2
I believe this is centripetal acceleration: $a_c=\frac{v^2}{r}=\omega^2r=\left(\frac{2\pi}{T}\r ight)^2r=\frac{4\pi^2r}{T^2}$.
Originally Posted by dabigchz
Unless your equation for centripetal force is somehow derived from ^ that one
So I had set gravitation force equal to centripetal force:
$\frac{GM_\oplus m}{r^2}=ma_c=\frac{4\pi^2rm}{T^2}$
$\frac{GM_\oplus}{r^2}=\frac{4\pi^2r}{T^2}$
$M_\oplus=\frac{4\pi^2r^3}{GT^2}$

Originally Posted by dabigchz
We were definitely supposed to use G.
As you can see, given all the other information, the $G$ is not necessary.