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Two persons of masses 75 kg and 80 kg sit 1 m apart. Taking $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, the gravitational force between them is:
A$4.002 \times 10^{-9}$ N
B$4.002 \times 10^{-5}$ N
C$4.002 \times 10^{-7}$ N
D$6.670 \times 10^{-7}$ N
Answer & Solution
Correct answer: C. $4.002 \times 10^{-7}$ N
1. Use Newton's law $F = \dfrac{G m_1 m_2}{r^2}$.
2. Substitute $G = 6.67\times10^{-11}$, $m_1 = 75$, $m_2 = 80$, $r = 1$.
3. $F = \dfrac{6.67\times10^{-11}\times75\times80}{1^2}$.
4. $75\times80 = 6000$, so $F = 6.67\times10^{-11}\times6000 = 4.002\times10^{-7}$ N.
5. The exponent is $-7$; values with $-5$ or $-9$ misplace the power of ten and are wrong.
_Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.15_
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