Question

State whether true or false:

$\log F=\log G+\log m_1+\log m_2-2\log d$ gives $F=G\frac{m_2{m}_1}{d^2}$.

• A. True
• B. False

Answer 1

The correct option is A True

$\log F=\log G+\log m_1+\log m_2-2\log d$

$\therefore \log F=\log G+\log m_1+\log m_2-\log d^2$ [Using ($\log a^b=b\log a$)]

$\therefore \log F=\log \frac{G{m}_1{m}_2}{d^2}$ [Using ($\log a.b=\log a+\log b,\log \frac{a}{b}=\log a-\log b$)]

$\therefore F=\frac{G{m}_1{m}_2}{d^2}$