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 r

Question

3 years ago

10

adius of Zeeman-474 is 6.38 × 10 6 m. What is the mass of this planet?

1 answer:

ehidna [41] · Answer 1 · 2022-06-03T20:24:06+03:00

Answer:

Mass of the planet = 6.0 × $10^{24}$

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 × $10^{6}$ + 230 × 10³ )³ / ( 6.67 × $10^{-11}$ ) × (89 × 60)²

= 4π² ( 6610000)³ / ( 6.67 × $10^{-11}$ ) × (89 × 60)²

= 5.99 × $10^{24}$

= 6.0 × $10^{24}$