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

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

Step-by-step 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 &= (G\, m_(a)\, m_(b))/(r^(2))\end{aligned}`.

In this question, it is given that
`G \approx 6.673 * 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 * 10^(-11)\; {\rm N}`.

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

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