Question

Initially a satellite of 100 kg is in a circular orbit of radius 1.5\(R_E\). This satellite can be moved to a circular orbit of radius 3\(R_E\) by supplying \(a \times 10^6\) J of energy. The value of a is_________ . (Take Radius of Earth \(R_E = 6 \times 10^6\) m and g = 10 m/s\(^2\)).

• A. 1000
• B. 150
• C. 100
• D. 500

Hint:

Remember that the total energy of a satellite in orbit is negative. Moving to a higher orbit (larger r) means the energy becomes less negative (i.e., it increases). Therefore, energy must be supplied to the satellite to move it to a higher orbit. The change in energy will be positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to calculate the energy required to move a satellite from a lower circular orbit to a higher circular orbit. This energy is the difference between the total mechanical energy of the satellite in the final and initial orbits.
Step 2: Key Formula or Approach:
The total mechanical energy of a satellite of mass `m` in a circular orbit of radius `r` around a central body of mass `M` is:
\[ E = -\frac{GMm}{2r} \] The energy required to change orbits is \( \Delta E = E_{\text{final}} - E_{\text{initial}} \).
We can express GM in terms of g and \(R_E\): \( g = \frac{GM}{R_E^2} \implies GM = gR_E^2 \).
So, the energy formula becomes:
\[ E = -\frac{gR_E^2 m}{2r} \] Step 3: Detailed Explanation:
1. Calculate Initial Energy (\(E_1\)):
Initial orbit radius \( r_1 = 1.5 R_E \).
\[ E_1 = -\frac{gR_E^2 m}{2(1.5 R_E)} = -\frac{gR_E m}{3} \] 2. Calculate Final Energy (\(E_2\)):
Final orbit radius \( r_2 = 3 R_E \).
\[ E_2 = -\frac{gR_E^2 m}{2(3 R_E)} = -\frac{gR_E m}{6} \] 3. Calculate Energy Supplied (\(\Delta E\)):
\[ \Delta E = E_2 - E_1 = \left(-\frac{gR_E m}{6}\right) - \left(-\frac{gR_E m}{3}\right) \] \[ \Delta E = gR_E m \left(-\frac{1}{6} + \frac{1}{3}\right) = gR_E m \left(-\frac{1}{6} + \frac{2}{6}\right) = \frac{gR_E m}{6} \] 4. Substitute the values:
m = 100 kg
g = 10 m/s\(^2\)
\(R_E = 6 \times 10^6\) m
\[ \Delta E = \frac{(10 \, \text{m/s}^2) \times (6 \times 10^6 \, \text{m}) \times (100 \, \text{kg})}{6} \] \[ \Delta E = 10 \times 10^6 \times 100 = 1000 \times 10^6 \, \text{J} \] Step 4: Final Answer:
The energy supplied is \( 1000 \times 10^6 \) J. Comparing this to the given expression \( a \times 10^6 \) J, we find that \( a = 1000 \). This corresponds to option (A).