Question

A sphere mass 40 kg is attracted by another sphere of mass 80 kg by force of 2.5 into 10 raise to minus 6 Newton when their centres are 30 millimeter apart find G ?
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Answer 1

GIVEN :

• Mass of sphere, m1= 40 kg.
• Mass of another sphere, m2 = 80 kg.
• Force, F = $\sf 2.5 \: \times \: 10^{-6} \: N.$
• Radius, R = 30 mm = 30/10 => 3 cm.

TO FIND :

• Find the gravitational force, G = ?

FORMULA USED :

• $\sf G \: = \: \dfrac {F \: \times \: R^{2}}{M_1 \: × \: M_2}$

SOLUTION :

Now, Using the formula of G,

$\tt G \: = \: \dfrac {F \: \times \: R^{2}}{M_1 \: × \: M_2}$

$\implies \sf G \: = \: \dfrac {2.5 \: \times \: 10^{-6} \: \times \: 3^{2}}{40 \: \times \: 80}$

$\implies \sf G \: = \: \dfrac {25 \: \times 9 \: \times \: 10^{-6}}{32 \: \times \: 1000}$

$\implies \sf G \: = \: \dfrac {225}{32} \: \times \: 10^{-9}$

$\implies \sf G \: = \: 7.03 \: \times \: 10^{-9} \: \dfrac {N.m^{2}}{kg^{2}}$

$\therefore \red {\boxed {\sf G \: = \: 7.03 \: \times \: 10^{-9} \: \dfrac {N.m^{2}}{kg^{2}}}}$