Question

# An asteroid, whose mass is 4.50×10-4 times the mass of Earth, revolves in a circular orbit...

An asteroid, whose mass is 4.50×10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 4 times the Earth's distance from the Sun. Calculate the period of revolution of the asteroid.

What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

here,

the mass of Asteroid , m = 4.5 * 10^-4 * me

the distance from the sun , R = 4 * Re

the time period , T = 2*pi*sqrt(R^3 /(G * M))

so, we can write

T / Te = 2*pi*sqrt(R^3 /(G * M)) /( 2*pi*sqrt(Re^3 /(G * M)))

T /Te = (R /Re)^3/2

T/1 yr = (4)^3/2

T = 8 yrs

the time period of asteroid is 8 yrs

the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth , R = KEa /KEe

R = 0.5 * m * v^2 /(0.5 * me * ve^2)

R = (m/me) * ( v/ve)^2

R = (m/me) * ( (2*pi*r /T) /(2*pi*re /Te))^2

R = (m/me) * ( (r * Te) /(re * T))^2

R = (4.5 * 10^-4) * ( (4 * (1/8)))^2

R = 1.125 * 10^-4