Answer :
The energy needed to move the satellite to a higher orbit is approximately [tex]\(2.94 \times 10^{10} \, \text{J}\)[/tex].
To determine the energy needed to move the satellite to a higher orbit, we can use the gravitational potential energy formula for an object in orbit:
[tex]\[ U = -\frac{GMm}{r} \][/tex]
where:
- U is the gravitational potential energy,
- G is the gravitational constant [tex](\(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\))[/tex],
- M is the mass of the Earth [tex](\(5.972 \times 10^{24} \, \text{kg}\))[/tex],
- m is the mass of the satellite [tex](\(970 \, \text{kg}\))[/tex],
- r is the distance from the center of the Earth to the satellite.
The initial orbital radius [tex](\(r_1\))[/tex] is the sum of the Earth's radius and the initial altitude:
[tex]\[ r_1 = R_{\text{Earth}} + h_1 \][/tex]
The final orbital radius [tex](\(r_2\))[/tex] is the sum of the Earth's radius and the final altitude:
[tex]\[ r_2 = R_{\text{Earth}} + h_2 \][/tex]
The change in potential energy [tex](\( \Delta U \))[/tex] is given by:
[tex]\[ \Delta U = U_2 - U_1 \][/tex]
Substitute the expressions for U and the values into the formula:
[tex]\[ \Delta U = -\frac{GMm}{r_2} + \frac{GMm}{r_1} \][/tex]
Now, calculate the values and solve for [tex]\( \Delta U \)[/tex]:
[tex]\Delta U[/tex] = [tex]-\frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)(5.972 \times 10^{24} \, \text{kg})(970 \, \text{kg})}{r_2} + \frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)(5.972 \times 10^{24} \, \text{kg})(970 \, \text{kg})}{r_1}[/tex]
Substitute the values of [tex]\(r_1\)[/tex] and [tex]\(r_2\)[/tex], which are [tex]\(R_{\text{Earth}} + h_1\)[/tex] and [tex]\(R_{\text{Earth}} + h_2\)[/tex], respectively.
Now, plug in the known values and calculate the change in potential energy [tex](\( \Delta U \))[/tex].
[tex]\[ \Delta U \approx 2.94 \times 10^{10} \, \text{J} \][/tex]
Therefore, the energy that must be added to the system to move the satellite into a circular orbit with an altitude of 205 km is approximately [tex]\(2.94 \times 10^{10} \, \text{J}\)[/tex].
