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
Orla Orla 3 years 2021-08-26T01:05:52+00:00 1 Answers 192 views 0

Answers ( )

  1. thienhuong
    0
    2021-08-26T01:07:23+00:00
    Reply

    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}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )