Question

1. a 6.3 kg bowling ball and a 7.1 kg bowling ball rest on a rack 0.85 m apart. (a) what is the force of gravity exerted on each of the balls by the other ball? (b) at what separation is the force of gravity between the balls equal to 2.0×10⁻⁹ n?

Explanation:

Step1: Identify the formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^2}$, where $G = 6.67\times10^{- 11}\ N\cdot m^2/kg^2$, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them.

Step2: Calculate the force in part (a)

Given $m_1 = 6.3\ kg$, $m_2 = 7.1\ kg$, and $r = 0.85\ m$.
Substitute the values into the formula:
$F=G\frac{m_1m_2}{r^2}=6.67\times 10^{-11}\frac{6.3\times7.1}{(0.85)^2}$
$F = 6.67\times10^{-11}\frac{44.73}{0.7225}$
$F=6.67\times10^{-11}\times61.91$
$F = 4.13\times10^{-9}\ N$

Step3: Solve for $r$ in part (b)

We know $F = 2.0\times10^{-9}\ N$, $m_1 = 6.3\ kg$, $m_2 = 7.1\ kg$, and $G = 6.67\times10^{-11}\ N\cdot m^2/kg^2$.
From $F = G\frac{m_1m_2}{r^2}$, we can re - arrange it to get $r^2=G\frac{m_1m_2}{F}$.
Substitute the values: $r^2 = 6.67\times10^{-11}\frac{6.3\times7.1}{2.0\times10^{-9}}$
$r^2=6.67\times10^{-11}\frac{44.73}{2.0\times10^{-9}}$
$r^2=6.67\times10^{-11}\times22365$
$r^2 = 0.149$
$r=\sqrt{0.149}\approx0.386\ m$

Answer:

(a) $4.13\times10^{-9}\ N$
(b) $0.386\ m$