Question

An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg. The center of a large ball is separated by 0.057 m from the center of a small ball. The Cavendish apparatus for measuring G. As the small spheres of mass m are attracted to the large spheres of mass M, the rod between the two small spheres rotates through a small angle. Find the magnitude of the gravitational force between the masses if the value of the universal gravitational constant is 6.67259 × 10−11 Nm2/kg2. Answer in units of N.

Answer 1

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$.

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$

So, the magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$. Hence, this is the required solution.