Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.