umanaisabel07 umanaisabel07
  • 22-12-2021
  • Physics
contestada

9) when the distance was one fourth as much, what happened to the force between the objects?

in this case (G=6.67E-11)

Respuesta :

leena
leena leena
  • 23-12-2021

Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = \frac{Gm_1m_2}{r^2}}[/tex]

G = Gravitational Constant

m1, m2 = mass of objects (kg)

r = distance between objects (m)

There is an INVERSE-SQUARE relationship between the gravitational force and the distance between the objects, so:

[tex]F_g = \frac{Gm_1m_2}{(\frac{1}{4}r)^2} = F_g = \frac{Gm_1m_2}{\frac{1}{16}r^2}[/tex]

[tex]= 16G\frac{m_1m_2}{r^2} = 16F_g[/tex]

Thus, the gravitational force between the objects would INCREASE by a factor of 16.

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