Question

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objects on a 69.0 kg object placed midway between them.

Answer 1

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$