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

    $W =  \frac{631.92  }{4.45}$

        $W  = 142 \ lb$