Question

Two balls, each with a mass of 0.803 kg,
exert a gravitational force of 8.27 × 10-¹¹ N
on each other.
11
How far apart are the balls? The value
of the universal gravitational constant is
6.673 x 10-¹¹ Nm²/kg².
Answer in units of m.
Answer in units of m

Answer 1

Answer:

Approximately $0.719\; {\rm m}$.

Explanation:

Let $G$ denote the constant of universal gravitation. Let $m_{a}$ and $m_{b}$ denote the mass of two objects. By Newton's Law of Universal Gravitation, if the distance between the two objects is $r$, the gravitational force of one object on the other will be:

$\begin{aligned}F &= \frac{G\, m_{a}\, m_{b}}{r^{2}}\end{aligned}$.

In this question, it is given that $G \approx 6.673 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}$. The mass of the two objects are $m_{a} = 0.803\; {\rm kg}$ and $m_{b} = 0.803\; {\rm kg}$. The gravitational force of one object on the other is $F = 8.27 \times 10^{-11}\; {\rm N}$.

Rearrange the equation to find the distance $r$ between these two objects:

$\begin{aligned}r &= \sqrt{\frac{G\, m_{a}\, m_{b}}{F}} \\ &\approx \sqrt{\frac{(6.673 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}})\, (0.803\; {\rm kg})\, (0.803\; {\rm kg})}{8.27 \times 10^{-11}\; {\rm N}}} \\ &= \sqrt{\frac{(6.673 \times 10^{-11})\, (0.803)\, (0.803)}{(8.27 \times 10^{-11})}}\; {\rm m} \\ &\approx 0.719\; {\rm m}\end{aligned}$.