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
A
6.67N
B
6.56N
C
7.5N
D
8.6N
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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 = \(6.67408\times10^{-11}Nm^2/kg^2\))
F = \(\frac{(6.67 \times 10^{– 11 + 24 + 27})}{ 10^{40}} = 6.67 \times \frac{10^{40}}{10^{40}}\)
= 6.67N
f = 6.67N
Source: Myschool Past Questions Archive (Record #40800)