Answer:
The workdone is [tex]W = 3.14*10^{10} \ J[/tex]
Explanation:
From the question we are told that
The equation for the force is [tex]F= \frac{G * m_1 * m_2}{r^2}[/tex]
The gravitational constant is [tex]G = 6.67*10^{-11} \ Nm^2/kg^2[/tex]
The mass of satellite is [tex]m_1 = 1400 \ kg[/tex]
The radius of the earth is [tex]r = 6.37*10^6 m[/tex]
The mass of earth is [tex]m_2 = 5.98* 10^ 24 \ kg[/tex]
Generally the work required is mathematically represented as
[tex]W = \int\limits^a_b {F} \, dr[/tex]
Where b = G this is because the satellite has to overcome the earth gravitational pull
And a = [tex]\infty[/tex] this because at infinity the there is not effect of gravitational force
So
[tex]W = \int\limits^{\infty}_{6*10^{-11}} {F} \, dr[/tex]
[tex]W = \int\limits^{\infty}_{6*10^{-11}} { \frac{G * m_1 * m_2}{r^2}} \, dr[/tex]
[tex]W = G * m_1 * m_2 \int\limits^{\infty}_{6*10^{-11}} { \frac{1}{r^2}} \, dr[/tex]
substituting values
[tex]W = 6.67*10^{-11} * 1400 *5.98*10^{24} \int\limits^{\infty}_{6*10^{-11}} { \frac{1}{r^2}} \, dr[/tex]
[tex]W = 2 *10^{17}\int\limits^{\infty}_{6*10^{-11}} { \frac{1}{r^2}} \, dr[/tex]
[tex]W = 2 *10^{17}} [{ - \frac{1}{r}} ] \left | \infty \atop {6*10^{-11}}} \right.[/tex]
[tex]W = 2 *10^{17}} [{ - \frac{1}{\infty }} + \frac{1}{6*10^{-11} }}][/tex]
[tex]W = 2 *10^{17}} [{ - 0 + \frac{1}{6*10^{-11} }}][/tex]
[tex]W = 2 *10^{17}} * 1.57*10^{-7}[/tex]
[tex]W = 3.14*10^{10} \ J[/tex]
