Question

Amazingly Chyna-Rose goes up another ramp and really turns her jetpack up to provide enough velocity to achieve orbit around the Earth! If the mass of the Earth is 5.97 x 1024 Kg and her period is 10.0 days, then determine Chyna-Rose’s orbital radius, orbital velocity, and centripetal acceleration.

Answer 1

Mass of the earth = M = 5.97 x 10²⁴ kg,

Time period of Chyna - Rose = T = 10 days = ( 10 x 24 x 3600 ) s,

R be Chyna-Rose’s orbital radius, v = 2R / T is its orbital velocity, and, m be its mass,

and, G = Universal gravitational constant.

The gravitational attraction force supplies the centripetal force of the rotation, i.e.,

GMm / R² = mv² / R

or, GM = ( 2R / T )² R = 4²R³ / T²

or, R³ = GMT² / 4² = 6.67 x 10^-11 x 5.97 x 10²⁴ x ( 10 x 24 x 3600 )² / ( 4 x 3.14² )

or, R³ ~ 7.537 x 10²⁴

or, R ~ 1.96 x 10⁸.

Hence, Chyna-Rose’s orbital radius is : 1.96 x 10⁸ m.

Hence, Chyna-Rose’s orbital velocity is :

v = 2R / T = 2 x 3.14 x 1.96 x 10⁸ m / ( 10 x 24 x 3600 ) s ~ 1424.63 m / s.

Hence, the centripetal acceleration of Chyna-Rose’s rotation is :

v² / R = ( 1424.63 m / s )² / ( 1.96 x 10⁸ m ) ~ 1.0355 x 10^-2 m / s².