Question

The value of the force of gravity on a boy of mass 40kg was found to be 2400N. Using ME=6×1024kg and Re=6.4×106m, find the value of G?
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Answer 1

Answer:-

$\orange{\bigstar}$ The value of G is

$\bf{4.096 \times 10^{-10} \: Nm^{2}/kg^{2}}$

• Given:-

Force of gravity on boy = 2400 N

Mass of the boy = 40 kg

$M_{e}$ = 6 × 10²⁴ kg

$R_{e}$ = 6.4 × 10^6 m

• To Find:-

Value of G

• Solution:-

We know,

$\boxed{\sf{F = \dfrac{G M_{e}m}{R_{e}^{2}}}}$

→ $\sf{G = \dfrac{F R_{e}^{2}}{M_{e}m}}$

→ $\sf{G= \dfrac{2400 (6.4 \times 10^6)^{2}}{6 \times 10^{24} \times 40}}$

→ $\sf{G = \dfrac{2400 \times 40.96 \times 10^{12}}{240 \times 10^{24}}}$

→ $\sf{G = \dfrac{40.96 \times 10^{13}}{10^{24}}}$

→ $\sf{G = 40.96 \times 10^{-11}}$

→ $\sf{G = 4.096 \times 10^{-10} \: Nm^{2}/kg^{2}}$

Therefore, the value of G is

$\bf{4.096 \times 10^{-10} \: Nm^{2}/kg^{2}}$

$\green{\bigstar}$Note:-

Universal accepted value of G is $\\ \bf{6.67 \times 10^{-11} \: Nm^{2}/kg^{2}}$