Question

From the gravitational law, calculate the weight W (gravitational force with respect to the earth) of a 70 kg spacecraft traveling in a circular orbit 275 km above the earth's surface. Express W in Newtons and pounds.

Answer 1

Answer:

The  value in Newton is $W = 631.92 \ N$

The  value in pounds is    $W = 142 \ lb$

Explanation:

From the question we are told that

  The  mass of the spacecraft is  $m = 70 \ kg$

   The distance above  the earth is  $d = 275 \ km = 275000 \ m$

Generally the gravitational force with respect to the earth is mathematically represented as

       $W = \frac{G * m * m_e}{ (d + r_e)^2}$

Here $m_e$ is the mass of earth with value $m_e = 5.978 *10^{24} \ kg$

       $r_e$ is the radius of the earth with value  $r_e = 6371 \ km = 6371000 \ m$

   G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3/ kg\cdot s^2$

So

     $W = \frac{ 6.67 *10^{-11} * 70 * 5.978 *10^{24}}{ (275000 + 6371000)^2}$

     $W = 631.92 \ N$

Converting to  pounds

    $W = \frac{631.92 }{4.45}$

        $W = 142 \ lb$