Question

The gravitational force of attraction between two masses is F

1. If both masses are tripled, what happens to the force?
2. If both masses are tripled, what happens to the force?
3. If masses are not changed, but the distance of separation is reduced to 2/3 the original distance, what happens to the force
4. If both masses are doubled and the distance of the separation is tripled, show what happens to the force
5. if one of the masses is doubled, the other remains unchanged, and the distance of separation is quadrupled, show what happens to the force.

6. What would be the value of Earth's gravitational field strength if Earth's mass was double its actual value, but its radius was reduced to one-third of its original value?

Answer 1

Hi there!

We can use the equation for gravitational force:
$F_g = \frac{Gm_1m_2}{r^2}$

Fg = force due to gravity (N)

m1, m2 = masses of objects (kg)

r = distance between the center of mass of the objects (m)

G = gravitational constant

1.

Using the equation:
$F'_g = \frac{G(3m_1)(3m_2)}{r^2} = 9F_g$

The force would be 9 times larger.

2. (Repeat?)

3.

The square of distance is inversely related, so:
$F'g= \frac{Gm_1m_2}{(\frac{2}{3}r)^2} = \frac{Gm_1m_2}{\frac{4}{9}r^2} = \frac{9}{4}F_g$

The force is 9/4ths times larger.

4.

Combining:
$F'_g = \frac{G(2m_1)(2m_2)}{(3r)^2} = \frac{4Gm_1m_2}{9r^2} = \frac{4}{9}F_g$

The force is 9/4ths times larger.

5.

$F'g = \frac{G(2m_1)m_2}{(4r)^2} = \frac{2Gm_1m_2}{16r^2} = \frac{1}{8}F_g$

The force is 1/8th the original.

6.

Equation for gravitational field:
$g = \frac{Gm_e}{r_e^2}$

g = acceleration due to gravity (m/s²)

G = gravitational constant

me = mass of earth (kg)
re = radius of earth (m)

With the following transformations:
$g' = \frac{G(2m_e)}{(\frac{1}{3}r)^2} = \frac{2Gm_e}{\frac{1}{9}r^2} = 18g$

The acceleration due to gravity is 18 times as large.