Question

The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 6.4 earth years. What are the asteroid's orbital radius and speed?

Answer 1

"Using Kepler's 3rd law :

T² = 4π²r³ / GM

find r :

r = [GMT² / 4π²]⅓

Where G =universal gravitational constant, M = mass of the sun,

T = asteroid's period in seconds

r = radius of the orbit.
Covert 6.4 years to seconds :

6.4 years = 6.4x365x24x60x60.
= 2.02 x 10^8 s

The radius of the orbit :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(2.02 x 10^8s)² / 4π²]⅓
= 5.16 x 10^11m

"The orbital speed can be found from the circular velocity equation : "

v = √[GM / r]
= √[(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg) / 5.16 x 10^11m]
= 1.6 x 10^4m/s "