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Thread: Tricky Physics Proof (that g=9.8)

  1. #1
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    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.
    Radius of moon's orbit is 60.1 x Earth's radius
    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.
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  2. #2
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    First, we set the centripetal acceleration equal to the gravitational force to get: $\displaystyle \omega^2=\frac{GM}{r^3}$. To relate this to a period, we note that $\displaystyle T=\frac{2\pi}{\omega}$, and solving for $\displaystyle M$ we get $\displaystyle M_\oplus=\frac{4\pi^2r^3}{GT^2}$. This gives us an expression for the mass of the earth given the period and radius of the moonís orbit.

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

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

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