Question

Newton measured the centripetal acceleration of the moon in its orbit around Earth by comparing the force Earth exerts on the moon with the force Earth exerts on an apple. He obtained a value of ac = 2.66×10-3 m/s2. If Newton had taken the mass of Earth to be ME = 5.97×1024 kg and the mean distance between the centers of Earth and the moon to be RME = 3.86×108 m, what value would he have obtained for the gravitational constant, in units of N⋅m2/kg2?

Answer 1

Answer:

$G = 6.64*10^{-11} Nm^2/kg^2$

Explanation:

According to Newton 2nd law of motion: F = ma where m is the mass of the moon and a is the centripetal acceleration of the gravitational force.

Also Gravitational force formula

$F_g = G\frac{m*ME}{S^2}$ where G is the gravitational force, ME is the mass of Earth and S is the mean distance between Earth and Moon.

As the moon is at constant circular motion around Earth, there forces balance out:

$F = F_g$

$ma = G\frac{m*ME}{S^2}$

$G = \frac{S^2*a}{ME}$

We can now substitute $S = RME = 3.86*10^8 m$ and $a = ac = 2.66*10^{-3} m/s^2$ and $ME = 5.97 * 10^{24} kg$

$G = \frac{(3.86*10^8)^2*2.66*10^{-3}}{5.97 * 10^{24}}$

$G = 6.64*10^{-11} Nm^2/kg^2$