Question

Two objects with masses, m1 and m2, are separated a distance d. The gravitational force between these masses is F. Suppose the object with a mass m1 is replaced by an object ten times the mass of m1. At the same time, the second object i also replaced.

If the distance and force between the two new objects remains the same, then what must be the new mass of the second object?
a. m2/10
b. 10²m2
c. m1−m2
d. 10m2
e. 10m1m2

Answer 1

`(m_(2) / 10)` (one-tenth the initial mass of this object.)

Step-by-step explanation:

The gravitational attraction between two point mass is:

`\displaystyle F = (G\, m_(1)\, m_(2))/(r^(2))`,

where:

• 
  `G` is the gravitational constant,
• 
  `m_(1)` and
  `m_(2)` are the mass of the two objects, and
• 
  `r` is the distance between the center of mass of the two objects.

In this question, it is given that:

• Mass of the first object is now
  `(10\, m_(1))`.
• The distance between the two objects
  `r` stays the same.

Let
`{m_(2)}^(\prime)` denote the new mass of the other object. The goal is to ensure that the gravitational force between the two objects stays the same. In other words:

`\displaystyle (G\, m_(1)\, m_(2))/(r^(2)) = \frac{G\, (10\, m_(1))\, ({m_(2)}^(\prime))}{r^(2)}`.

Rearrange this equation to find
`{m_(2)}^(\prime)`:

`\begin{aligned}{m_(2)}^(\prime) &= (\displaystyle (G\, m_(1)\, m_(2))/(r^(2)))/(\displaystyle (G\, (10\, m_(1)))/(r^(2))) = (m_(2))/(10)\end{aligned}`.

In other words, the new mass of the other object should be
`(m_(2) / 10)`, which is one-tenth the initial value.