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

`F_(net) = 6.879* 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* 10^(-11)* 120* 0.69}{\frac({0.380}{2}^(2))} = 1.529* 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* 10^(-11)* 420* 0.69}{\frac({0.380}{2}^(2))} = 5.35* 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* 10^(- 7) + 5.35* 10^(- 7) = 6.879* 10^(- 7)\ N`