A 9.0 × 10 3 kg satellite orbits the earth at the distance of 2.56 × 10 7 m from Earth's surface. What is its period?

1.6537 * 10^9

Explanation:

Given that :

Distance (r) = 2.56 × 10^7 m

Using the relation from Kepler's third law of motion :

T2 = [ 4* pi^2/ G (M1 + M2) ]r^3

M1 = mass of earth = 5.97 *10^24

M2 = Mass of satellite = 9.0 × 10^3 kg

G = Gravitational constant = 6.67 *10^-11

T2 = [ (4 x 3.14^2) / 6.67 x 10^-11 ( 9 x 10^3 + 5.97 x 10^24)]/ (2.56 x 10^7)^3

= 1.6537 * 10^9

unlike poles attract

overcharge is the right answer system