Hi there!
We can use the equation for gravitational force:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
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:
[tex]F'_g = \frac{G(3m_1)(3m_2)}{r^2} = 9F_g[/tex]
The force would be 9 times larger.
2. (Repeat?)
3.
The square of distance is inversely related, so:
[tex]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[/tex]
The force is 9/4ths times larger.
4.
Combining:
[tex]F'_g = \frac{G(2m_1)(2m_2)}{(3r)^2} = \frac{4Gm_1m_2}{9r^2} = \frac{4}{9}F_g[/tex]
The force is 9/4ths times larger.
5.
[tex]F'g = \frac{G(2m_1)m_2}{(4r)^2} = \frac{2Gm_1m_2}{16r^2} = \frac{1}{8}F_g[/tex]
The force is 1/8th the original.
6.
Equation for gravitational field:
[tex]g = \frac{Gm_e}{r_e^2}[/tex]
g = acceleration due to gravity (m/s²)
G = gravitational constant
me = mass of earth (kg)
re = radius of earth (m)
With the following transformations:
[tex]g' = \frac{G(2m_e)}{(\frac{1}{3}r)^2} = \frac{2Gm_e}{\frac{1}{9}r^2} = 18g[/tex]
The acceleration due to gravity is 18 times as large.
