Question:
A 1200 kg artificial satellite is in an orbit of radius $2R_{E}$ about the earth. The energy required to transfer it to an orbit of radius $3R_{E}$ is ($g=10\text{ms}^{-2}$, $R_{E}=6400 \text{ km}$)
A 1200 kg artificial satellite is in an orbit of radius $2R_{E}$ about the earth. The energy required to transfer it to an orbit of radius $3R_{E}$ is ($g=10\text{ms}^{-2}$, $R_{E}=6400 \text{ km}$)
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Remember that orbital shift energy $\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. Substitute $GM = gR_E^2$ for easy numerical calculations.
Updated On: Jun 3, 2026
- $1.2\times10^{9} \text{ J}$
- $6.4\times10^{9} \text{ J}$
- $6400\times10^{3} \text{ J}$
- $3.2\times10^{9} \text{ J}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The total mechanical energy $E$ of a satellite revolving around the Earth in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$, where $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $G$ is the gravitational constant.
Step 2: Meaning
Since $g = \frac{GM}{R_E^2}$, we can substitute $GM = gR_E^2$ into the energy formula to obtain $E = -\frac{mgR_E^2}{2r}$.
Step 3: Analysis
The initial energy in the orbit of radius $r_1 = 2R_E$ is $E_1 = -\frac{mgR_E^2}{4R_E} = -\frac{mgR_E}{4}$. The final energy in the orbit of radius $r_2 = 3R_E$ is $E_2 = -\frac{mgR_E^2}{6R_E} = -\frac{mgR_E}{6}$. The required energy to change the orbit is $\Delta E = E_2 - E_1 = mgR_E \left(\frac{1}{4} - \frac{1}{6}\right) = \frac{mgR_E}{12}$. Substituting the given values: $m = 1200 \text{ kg}$, $g = 10 \text{ ms}^{-2}$, and $R_E = 6.4 \times 10^6 \text{ m}$, we get $\Delta E = \frac{1200 \times 10 \times 6.4 \times 10^6}{12} = 100 \times 10 \times 6.4 \times 10^6 = 6.4 \times 10^9 \text{ J}$.
Step 4: Conclusion
Thus, the total external work or energy required for the orbital shifting is $6.4 \times 10^9 \text{ J}$.
Final Answer: (B)
The total mechanical energy $E$ of a satellite revolving around the Earth in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$, where $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $G$ is the gravitational constant.
Step 2: Meaning
Since $g = \frac{GM}{R_E^2}$, we can substitute $GM = gR_E^2$ into the energy formula to obtain $E = -\frac{mgR_E^2}{2r}$.
Step 3: Analysis
The initial energy in the orbit of radius $r_1 = 2R_E$ is $E_1 = -\frac{mgR_E^2}{4R_E} = -\frac{mgR_E}{4}$. The final energy in the orbit of radius $r_2 = 3R_E$ is $E_2 = -\frac{mgR_E^2}{6R_E} = -\frac{mgR_E}{6}$. The required energy to change the orbit is $\Delta E = E_2 - E_1 = mgR_E \left(\frac{1}{4} - \frac{1}{6}\right) = \frac{mgR_E}{12}$. Substituting the given values: $m = 1200 \text{ kg}$, $g = 10 \text{ ms}^{-2}$, and $R_E = 6.4 \times 10^6 \text{ m}$, we get $\Delta E = \frac{1200 \times 10 \times 6.4 \times 10^6}{12} = 100 \times 10 \times 6.4 \times 10^6 = 6.4 \times 10^9 \text{ J}$.
Step 4: Conclusion
Thus, the total external work or energy required for the orbital shifting is $6.4 \times 10^9 \text{ J}$.
Final Answer: (B)
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