1) The new gravitational force is half of the original force
2) The new gravitational force is 1/8 of the original force
Explanation:
1)
Let's call F the initial gravitational force between the object A and B. The magnitude of F is given by the equation
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
are the masses of the two objects
R is the separation between them
In this part of the problem, we are told that the distance between the two objects doubles, so the new distance is
R' = 2R
While the mass of object A is doubled, so the new mass is
![m_A' = 2m_A](https://tex.z-dn.net/?f=m_A%27%20%3D%202m_A)
Therefore, the new gravitational force is
![F' = \frac{Gm_A' m_B}{R'^2}= \frac{G(2m_A) m_B}{(2R)^2}=\frac{1}{2}(\frac{Gm_A m_B}{R^2})=\frac{F}{2}](https://tex.z-dn.net/?f=F%27%20%3D%20%5Cfrac%7BGm_A%27%20m_B%7D%7BR%27%5E2%7D%3D%20%5Cfrac%7BG%282m_A%29%20m_B%7D%7B%282R%29%5E2%7D%3D%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7BGm_A%20m_B%7D%7BR%5E2%7D%29%3D%5Cfrac%7BF%7D%7B2%7D)
Therefore, the new force is half of the original force.
2)
In this second part of the problem, the distance between the two objects doubles, so the new distance is
R' = 2R
While the mass of object A is halved, so the new mass is
![m_A' = \frac{m_A}{2}](https://tex.z-dn.net/?f=m_A%27%20%3D%20%5Cfrac%7Bm_A%7D%7B2%7D)
Therefore, the new gravitational force this time is:
![F' = \frac{Gm_A' m_B}{R'^2}= \frac{G(m_A/2) m_B}{(2R)^2}=\frac{1}{8}(\frac{Gm_A m_B}{R^2})=\frac{F}{8}](https://tex.z-dn.net/?f=F%27%20%3D%20%5Cfrac%7BGm_A%27%20m_B%7D%7BR%27%5E2%7D%3D%20%5Cfrac%7BG%28m_A%2F2%29%20m_B%7D%7B%282R%29%5E2%7D%3D%5Cfrac%7B1%7D%7B8%7D%28%5Cfrac%7BGm_A%20m_B%7D%7BR%5E2%7D%29%3D%5Cfrac%7BF%7D%7B8%7D)
Therefore, the new force is 1/8 of the original force.
Learn more about gravitational force:
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brainly.com/question/12785992
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