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A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 20.0 m/s2, what is the radius of the planet?

User Maarty
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Answer:

Step-by-step explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² =
(4* \pi^2* r^3)/(GM) where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

User ZAIRI Oussama
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