Calculate the gravitation force of attraction between two planets of \(10^{24}\)kg and \(10^{27}\)kg separated by a distance of \(10^{20}\)meters
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6.67N
- 6.56N
- 7.5N
- 8.6N
The correct answer is: A
Explanation
F = \(\frac{Gm_1 m_2}{r^2}\)
\(M_1 = 10^{24}\)Kg, \(M_2 = 10^{27}\)Kg , r = \(10^{20}\)meters
(Note: the gravitational constant of G = \(6.67408×10^{-11}Nm^2/kg^2\))
F = \(\frac{(6.67 \times 10^{– 11 + 24 + 27})}{ 10^{40}} = 6.67 \times \frac{10^{40}}{10^{40}}\)
= 6.67N
f = 6.67N
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