Question

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences due to the person's gravitational pull?

Answer 1

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$