Question

The force of attraction between two spheres of masses 40 kg and 15 kg is 9.8 into 10 ^ - 7 Newton when their centres are 0.2 metre apart find the value of capital G

Answer 1

Answer:

Given,

               The masses of the two spheres ($m_{1}$ & $m_{2}$) are 40 Kg & 15 Kg.

               Force of attraction between the spheres (F) 9.8 × $10^{-7}$ Newton

               Distance between the centers, r = 0.2 Meters

We know that

   The universal force of attraction, F = G $\frac{m_{1} m_{2}}{r^{2} }$

Therefore,                      G = (F $r^{2}$)/($m_{1} m_{2}$)

Putting the values we get,

                                       G = ( 9.8 ×$10^{-7}$ × $(0.2)^{2}$) / (40 × 15)

                                           = 6.5 × $10^{-11}$ N$m^{2}$/$kg^{2}$

The required value of G is 6.5 × $10^{11}$ N$m^{2}$/ $kg^{2}$

Explanation:

We know that

   The universal force of attraction, F = G $\frac{m_{1} m_{2}}{r^{2} }$

Therefore,                      G = (F $r^{2}$)/($m_{1} m_{2}$)

Putting the values in the above equation, we get the required answers for such problems.

Note: G is known as Universal Gravitational Constant which has a constant value of 6.647 ×$10^{-11}$ N$m^{2} / kg^{2}$. We get the answer close to this value because of round off in the given values.