Question

The gravitational force of attraction between two masses is F

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.