Answer to Question #96385 in Mechanics | Relativity for A

Question

A 5.4kg rock and a 8.8x10^–4 kg pebble are held near the surface of the earth. (A) determine the magnitude of the gravitational force exerted on each by the earth. (B) calculate the magnitude of the acceleration of each object when released.

Answer 1

According to the universal law of gravity, the magnitude of the gravitational force exerted on an object near the surface of Earth is $F = G \frac{m M_e}{R_e^2}$, where $M_e = 5.97 \cdot 10^{24} kg, R_e = 6.38 \cdot 10^6 m$ - mass and radius of Earth respectively, $G = 6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ - gravitational constant, $m$ - mass of the object. Hence,

A) $F_{rock} = 52.83 N$, $F_{pebble} = 8.6 \cdot 10^{-3} N$.

B) The magnitude of acceleration is $\frac{F}{m} = \frac{G M_e}{R_e^2} = g = 9.81 \frac{m}{s^2}$ (gravity of Earth) is independent of the mass of the object, and hence the same for both rock and pebble.