If Earth has a mass of 5.972 × 10^24 kg and Mars has a mass of 6.39 × 10^23 kg, and the distance between them is 225 x 10^6 m, with what force does Earth pull on Mars?

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If Earth has a mass of 5.972 × 10^24 kg and Mars has a mass of 6.39 × 10^23 kg, and the distance between them is 225 x 10^6 m, with what force does Earth pull on Mars?

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Answer 1 · 2018-03-06T22:37:11

This is an application of the Universal Gravitation Equation $F_{G} = G \frac{m_{1} m_{2}}{d^{2}}$ $F_G=G{m_1m_2}/d^2$

Explanation:

In your case, you are given all the variable and all you need to know is that the Gravitational Constant, $G = 6.67 \times 10^{- 11} \frac{m^{3}}{k g \cdot s^{2}}$ $G = 6.67xx10^{-11} {m^3}/{kg*s^2}$

So, $F_{G} = 6.67 \times 10^{- 11} \frac{5.972 \times 10^{24} \cdot 6.3 \times 10^{23}}{{(225 \times 10^{6})}^{2}}$ $F_G=6.67xx10^{-11}{5.972xx10^24*6.3xx10^23}/{(225xx10^6)^2}$

Then, $F_{G} = 6.67 \times 10^{- 11} \frac{3.82 \times 10^{48}}{(50625 \times 10^{12})}$ $F_G=6.67xx10^{-11}{3.82xx10^48}/{(50625xx10^12)}$

$F_{G} = 6.67 \times 10^{- 11} \frac{3.82 \times 10^{48}}{(5.06 \times 10^{16})}$ $F_G=6.67xx10^{-11}{3.82xx10^48}/{(5.06xx10^16)}$

$F_{G} = 6.67 \times 10^{- 11} \cdot 7.55 \times 10^{31}$ $F_G=6.67xx10^{-11}*7.55xx10^31$

$F_{G} = 5.04 \times 10^{21}$ $F_G=5.04xx10^21$ Newtons

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If Earth has a mass of 5.972 × 10^24 kg and Mars has a mass of 6.39 × 10^23 kg, and the distance between them is 225 x 10^6 m, with what force does Earth pull on Mars?

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