Question

An asteroid, whose mass is 3.6 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 1.6 times the Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

Answer 1

Answer:

a) 2.02 years

b) 8.1 x 10⁻⁸.

Explanation:

Time period of a rotating body T² is proportional to radius of orbit R³ So

T₁² / T₂² = R₁³ /R₂³ ( T₁ and R₁ is time period and radius of orbit of the earth .)

1² / T₂² =( 1/1.6)³

T₂ = 2.02 years.

Kinetic energy of an orbiting body = 1/2 m v₀² ( v₀ is orbital speed)

= 1/2 m x 2 g R = m x G m/R² X R= m² x G /R

Kinetic energy of asteroid K₁ / kinetic energy of earth K₂ =

(mass of asteroid/mass of earth)² x( radius of earth / radius of asteroid)

=( 3.6 x 10⁻⁴)² x 1/1.6 = 8.1 x 10⁻⁸ .