Question

if sphere of mass 45 kg is placed at a distance of 30cm from another sphere of mass 75kg. the two sphere mutually attract each other with a gravitational force equal to weight of 1/4th of a milligram. compute the value of the gravitational constant

Answer 1

Given:

Sphere of mass 45 kg is placed at a distance of 30cm from another sphere of mass 75kg. the two sphere mutually attract each other with a gravitational force equal to weight of 1/4th of a milligram.

To find:

Value of Gravitational Constant.

Calculation:

According to NEWTON'S LAW OF GRAVITATION:

$\therefore \: force = \dfrac{G(m1)(m2)}{ {r}^{2} }$

$= > \: \dfrac{ {10}^{ - 3} g}{4} = \dfrac{G(45)(75)}{ {( \frac{30}{100}) }^{2} }$

$= > \: \dfrac{ {10}^{ - 3} \times 10}{4} = \dfrac{G(45)(75)}{ {( \frac{30}{100}) }^{2} }$

$= > \: \dfrac{ {10}^{ - 2}}{4} = \dfrac{G(45)(75)}{ {( \frac{30}{100}) }^{2} }$

$= > \: \dfrac{ {10}^{ - 2}}{4} = \dfrac{G(45)(75)}{ {( \frac{3}{10}) }^{2} }$

$= > \: \dfrac{ {10}^{ - 2}}{4} = \dfrac{G(45)(75)}{ \frac{9}{100}}$

$= > \: \dfrac{ {10}^{ - 2}}{4} = \dfrac{G(45)(75) \times 100}{9}$

$= > \: G = \dfrac{ {10}^{ - 2} \times 9}{4 \times 45 \times 75 \times 100}$

$= > \: G = \dfrac{ {10}^{ - 4} \times 9}{4 \times 45 \times 75 }$

$= > \: G = 0.00066 \times {10}^{ - 4}$

$= > \: G = 6.6 \times {10}^{ - 8} \: N {m}^{2} {kg}^{ - 2}$

So, the final value comes as:

$\boxed{ \bold{\: G = 6.6 \times {10}^{ - 8} \: N {m}^{2} {kg}^{ - 2} }}$