91Ó°ÊÓ

First, we need to calculate the gravitational potential energy (GPE) required to lift the satellite from the surface of the Earth to a height of 800 km. The GPE at a height \( h \) is given by:\[U = G \frac{mM}{R+h} - G \frac{mM}{R} = m g R \left(1 - \frac{R}{R + h}\right)\]Where:- \( m = 2000 \; \text{kg} \) is the mass of the satellite,- \( g = 10 \; \text{m/s}^2 \) is the acceleration due to gravity at the surface,- \( R = 6400 \times 10^3 \; \text{m} \) is the radius of the Earth,- \( h = 800 \times 10^3 \; \text{m} \) is the altitude.Now, substitute the values:\[U = 2000 \times 10 \times 6400 \times 10^3 \times \left(1 - \frac{6400}{7200}\right)\]Calculating this gives:\[U \approx 2000 \times 10 \times 6400 \times 10^3 \times \frac{800}{7200} = 2000 \times 10 \times 6400 \times 10^3 \times 0.1111 \approx 14.22 \times 10^{10} \; \text{J}\]

Next, we calculate the kinetic energy (KE) required for the satellite to maintain a circular orbit at 800 km above the Earth's surface. The formula for KE in orbit is:\[KE = \frac{1}{2} \cdot m \cdot v^2\]where orbital velocity \( v \) is given by:\[v = \sqrt{\frac{GM}{R+h}}\]Given that:- \( gR^2 = GM \), we substitute \( GM \) to find \( v \):\[v = \sqrt{gR^2 / (R + h)}\]Substitute values:\[v = \sqrt{10 \times 6400^2 / 7200} \approx 7358 \; \text{m/s}\]Now calculate \( KE \):\[KE = \frac{1}{2} \times 2000 \times (7358)^2 \approx 54.1 \times 10^{10} \; \text{J}\]

To find the total energy required, sum the gravitational potential energy and kinetic energy needed:\[\text{Total Energy} = U + KE \approx 14.22 \times 10^{10} + 54.1 \times 10^{10} = 68.32 \times 10^{10} \; \text{J}\]

Since the problem asks for energy in terms of \( \times 10^{10} \; \text{J} \), the total energy required is 68.32 \( \times 10^{10} \) J as calculated in the previous step.