A satellite of mass 100 kg is placed initially in a temporary orbit 80...
Consider a satellite of mass m is orbiting around the earth with orbit velocity n. Let h be the height of satellite from the surface of earth and r be the orbital radius of the satellite. Then
Total energy the orbiting satellite (E)
Work done in shifting the satellite from height h to h
1 is
W = difference in energy of satellite in two orbits
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A satellite of mass 100 kg is placed initially in a temporary orbit 80...
Given:
Mass of the satellite, m = 100 kg
Initial orbit radius, r1 = 800 km + 6400 km = 7200 km
Final orbit radius, r2 = 2000 km + 6400 km = 8400 km
Gravitational acceleration, g = 10 m/s^2
To find:
Amount of work done to move the satellite from the temporary orbit to the permanent orbit
Solution:
Step 1: Calculate the gravitational potential energy in the temporary orbit.
Gravitational potential energy (U1) is given by the formula:
U1 = -GMm/r1
where G is the universal gravitational constant (6.674 × 10^-11 N m^2/kg^2) and M is the mass of the Earth (5.972 × 10^24 kg).
Converting the radius to meters:
r1 = 7200 km = 7200 × 10^3 m
Plugging the values into the formula:
U1 = - (6.674 × 10^-11 N m^2/kg^2) × (5.972 × 10^24 kg) × (100 kg) / (7200 × 10^3 m)
U1 = - 4.906 × 10^8 J
Step 2: Calculate the gravitational potential energy in the permanent orbit.
Using the same formula, but with the new radius:
r2 = 8400 km = 8400 × 10^3 m
Plugging the values into the formula:
U2 = - (6.674 × 10^-11 N m^2/kg^2) × (5.972 × 10^24 kg) × (100 kg) / (8400 × 10^3 m)
U2 = - 2.668 × 10^8 J
Step 3: Calculate the work done.
The work done (W) is given by the formula:
W = U2 - U1
Plugging in the values:
W = - 2.668 × 10^8 J - (- 4.906 × 10^8 J)
W = 2.238 × 10^8 J
Converting to mega joules:
W = 2.238 × 10^8 J / (10^6 J/MJ)
W = 223.8 MJ ≈ 224 MJ
Therefore, the amount of work done to move the satellite from the temporary orbit to the permanent orbit is approximately 224 mega joules.