Question

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × $10^{24}$ kg and radius of the earth is 6.4 × $10^6$ m.)

Answer 1

According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by

F = $\frac{GMm}{r^2}$
Where,
Mass of Earth, M = 6 × $10^{24}$ kg
Mass of object, m = 1 kg
Universal gravitational constant, G = 6.7 × $10^{−11} Nm^2 kg^{−2 }$
Since the object is on the surface of the Earth,
$r$ = radius of the Earth ($R$)
$r$ = $R$ = 6.4 × $10^6$ m

Therefore, the gravitational force
$F$ = $\frac{GMm}{r^2}$

$F$ = $\frac{6.7×10^{−11}× 6×10^{24}×1}{ (6.4×10^6)^2}$

$F$ = 9.8 𝑁