Question

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. The Moon has a mass of 7.36 × 1022 kg; the Earth has a mass of 5.98 × 1024 kg; and the Sun has a mass of 1.99 × 1030 kg. The distance between the Moon and the Earth is 3.84 × 108 m; the distance between the Earth and the Sun is 1.496 × 1011 m. Calculate the total gravitational potential energy for this arrangement.

Answer 1

Answer:

U = 5.37*10^33 J

Explanation:

The gravitational potential energy between two bodies is given by:

$U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}$

G: Cavendish's constant = 6.67*10^-11 m^3/kg.s

For three bodies the total gravitational potential energy is:

$U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}]$

BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:

$U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J$

hence, the total gravitational energy is 5.37*10^33 J