Question

if your mass is 72 kg, your textbooks mass is 3.7 kg, and you and your textbook are separated by a distance of 0.33 m, what is the gravitational force between you and your textbook? newtons law of gravitation is $f_{gravity}=\frac{gm_1m_2}{r^2}$. the gravitational constant g is $6.67\times10^{-11}n\cdot m^2/kg^2$.

a. $1.63\times10^{-7}n$
b. $4.94\times10^{-7}n$
c. $2.45\times10^{3}n$
d. $5.38\times10^{-8}n$

Explanation:

Step1: Substitute values into formula

Given \(G = 6.67\times10^{-11}\space N\cdot m^{2}/kg^{2}\), \(m_1 = 72\space kg\), \(m_2 = 3.7\space kg\), \(r = 0.33\space m\).
\(F_{gravity}=\frac{Gm_1m_2}{r^{2}}=\frac{6.67\times 10^{-11}\times72\times3.7}{(0.33)^{2}}\)

Step2: Calculate numerator and denominator

First calculate numerator: \(6.67\times 10^{-11}\times72\times3.7=(6.67\times72\times3.7)\times10^{-11}\)
\(6.67\times72 = 480.24\), \(480.24\times3.7=1776.888\)
Denominator: \((0.33)^{2}=0.1089\)
So \(F_{gravity}=\frac{1776.888\times 10^{-11}}{0.1089}\)

Step3: Final calculation

\(\frac{1776.888}{0.1089}\approx16316.69\)
\(F_{gravity}=16316.69\times10^{-11}=1.63\times10^{-7}\space N\)

Answer:

A. \(1.63\times 10^{-7}\space N\)