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    Suppose the earth and the moon became equally charged by a sudden flow of protons from the sun. How much charge would need to be gained by each body to enable the electrostatic force of repulsion to overcome the gravitational force of attraction between them. The mass of the earth is 6.0x10 (to the power of 24) kg and the mass I'd the Moon is 6.1x 10 (to the power of 22) kg

    Answer: 5.23x 10 (to the power of 13) C
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    (Original post by 99ja99)
    Suppose the earth and the moon became equally charged by a sudden flow of protons from the sun. How much charge would need to be gained by each body to enable the electrostatic force of repulsion to overcome the gravitational force of attraction between them. The mass of the earth is 6.0x10 (to the power of 24) kg and the mass I'd the Moon is 6.1x 10 (to the power of 22) kg

    Answer: 5.23x 10 (to the power of 13) C
    My answer is 5.21 x 10 ^ 13 ... This was how I did it :

    Gravitational Force = GMm/(R^2), where G is 6.67 x 10 ^ -11
    Electrostatic Force = Qq/(4pi x epsilon x R^2) , where epsilon = 8.85 x 10 ^ -12

    Gravitational Force = Electrostatic Force (this will allow us to find the minimum charge of each body such that the electrostatic force is able to overcome the gravitational force between the bodies)

    GMm/(R^2) = Qq/(4pi x epsilon x R^2) , R^2 cancels out
    GMm = Qq/(4pi x epsilon)
    GMm x 4pi x epsilon = Qq, Both bodies are equally charged, so Qq = Q^2

    Sub in the values and square root GMm x 4pi x epsilon and you'll get the answers. I can't make the equation look cleaner :/ I tried.
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    (Original post by AndyLai9879)
    My answer is 5.21 x 10 ^ 13 ... This was how I did it :

    Gravitational Force = GMm/(R^2), where G is 6.67 x 10 ^ -11
    Electrostatic Force = Qq/(4pi x epsilon x R^2) , where epsilon = 8.85 x 10 ^ -12

    Gravitational Force = Electrostatic Force (this will allow us to find the minimum charge of each body such that the electrostatic force is able to overcome the gravitational force between the bodies)

    GMm/(R^2) = Qq/(4pi x epsilon x R^2) , R^2 cancels out
    GMm = Qq/(4pi x epsilon)
    GMm x 4pi x epsilon = Qq, Both bodies are equally charged, so Qq = Q^2

    Sub in the values and square root GMm x 4pi x epsilon and you'll get the answers. I can't make the equation look cleaner :/ I tried.
    Good clear explanation, although we prefer to give hints in the forum rather than full solutions immediately, to help the OP to reach the answer by themselves. Something to bare in mind in future...

    You could use  \LaTeX to make your equations look neater. It's built in to The Student Room, and enabled by typing between the tags [.tex] and [./tex], without the "." .

    You can find the guide to  \LaTeX here: https://www.thestudentroom.co.uk/help/latex

    \dfrac{GM_1M_2}{R^2} = \dfrac{Q_1Q_2}{4 \pi \epsilon_0 R^2}, etc...

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    (Original post by K-Man_PhysCheM)
    Good clear explanation, although we prefer to give hints in the forum rather than full solutions immediately, to help the OP to reach the answer by themselves. Something to bare in mind in future...

    You could use  \LaTeX to make your equations look neater. It's built in to The Student Room, and enabled by typing between the tags [.tex] and [./tex], without the "." .

    You can find the guide to  \LaTeX here: https://www.thestudentroom.co.uk/help/latex

    \dfrac{GM_1M_2}{R^2} = \dfrac{Q_1Q_2}{4 \pi \epsilon_0 R^2}, etc...

    Ah I see, didn't knew about that. Sorry!

    Also thanks for the guide to the LaTeX
 
 
 
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