Answer:
WT = 3.32*10^34 J
Explanation:
The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:
[tex]W_T=\int F_g dr[/tex] (1)
Fg: gravitational force between Sun and Earth
The gravitational force is given by:
[tex]F_g=G\frac{m_sm_e}{r^2}[/tex] (2)
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
ms: mass of the sun = 1.989*10^30 kg
me: mass of the Earth = 5.972 × 10^24 kg
r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km
You replace the formula (2) in (1):
[tex]W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}[/tex]
Next, you replace the values of all variables and solve obtain WT:
[tex]W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J[/tex]
hence, the work done on the Earth, in one orbit, is 3.32*10^34 J
