Question

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satellite's orbital speed.(b) Find the period of its revolution. (c) Find the gravitational force acting on it.

Answer 1

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = $\dfrac{mv^2}{r}$

     gravitational force = $\dfrac{GMm}{r^2}$

    equating both the above equation

    $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

      $v = \sqrt{\dfrac{GM}{r}}$

      $v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

          v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

   $T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

   $T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

          T = 14416.92 s

          $T = \dfrac{14416.92}{3600}\ hr$

          T = 4 hr

c) gravitational force acting

  $F = \dfrac{GMm}{r^2}$

  $F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

     F = 5202 N