Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
[tex]F = G\frac{m_1m_2}{R^2}
The problem states that the distance decrease to 1/10 of the original distance, this means:
[tex]R_2 = \frac{1}{10}R[/tex]
And the force at this distance would be written in terms of the same equation:
[tex]F_2 = G\frac{m_1m_2}{R_2^2}[/tex]
Find the ratio between the final and the initial force:
[tex]\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}[/tex]
Substitute the value for the final distance in terms of the initial distance:
[tex]\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}[/tex]
Simplify:
[tex]\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}[/tex]
This means the new force will be \frac{1}{100} of the original force.
