Share
A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The radius of Zee
Question
A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The radius of Zeeman-474 is 6.38 × 10 6 m. What is the mass of this planet?
in progress
0
Physics
3 years
2021-08-26T01:05:52+00:00
2021-08-26T01:05:52+00:00 1 Answers
192 views
0
Answers ( )
Answer:
Mass of the planet = 6.0 ×
Explanation:
Time period = 2π (R + h) / v
Orbital speed (v) = √GM / (R + h)
T² = 4π² (R + h)² / (GM/ (R + h))
= 4π² (R + h)³ / GM
making m the subject of the formula
m = 4π² (R + h)³ / GT²
= 4π² ( 6.38 × + 230 × 10³ )³ / ( 6.67 × ) × (89 × 60)²
= 4π² ( 6610000)³ / ( 6.67 × ) × (89 × 60)²
= 5.99 ×
= 6.0 ×