Physics - Help requested.

Question

Assume that a satellite orbits earth 230,000 m above its surface. given that the mass of earth is 5,97 x 10^24 kg, and the radius of earth is 6.38 x 10^6 m what is the satellite's period?

Answer 1

Answered this exact question yesterday 

Answer 2

Use Kepler's third law.

 

T² / r³ = 4π² / GM

 

T = √(4π²r³ / GM)

 

r = h + R_E = (230,000 + 6.38 x 10⁶) m = 6.61 x 10⁶ m

 

T = √[4π²(6.61 x 10⁶)³ / (6.67 x 10^-11)(5.97 x 10²⁴)] ≅ 5351 seconds