KCET · Physics · Gravitation
A satellite is orbiting close to the earth and has a kinetic energy \( \mathrm{K} \). The minimum extra kinetic
energy required by it to just overcome the gravitation pull of the earth is
- A \( \sqrt{3} \mathrm{~K} \)
- B \( \mathrm{K} \)
- C \( 2 \sqrt{2} K \)
- D - \( 2 \mathrm{~K} \)
Answer & Solution
Correct Answer
(B) \( \mathrm{K} \)
Step-by-step Solution
Detailed explanation
(B)
We know \( v_{e}=\sqrt{2} v_{0} \)
When satellite is in orbit \( K=\frac{1}{2} m v_{0}{ }^{2} \)
For satellite to escape its velocity should be \( v_{e} \)
\( \therefore K E \) To escape, \( K_{e}=\frac{1}{2} m v_{e}^{2}=\frac{1}{2} m\left(\sqrt{2} v_{0}\right)^{2}=2 K \)
\( \therefore \) Extra \( K . E \). required \( =2 K-K=K \)
We know \( v_{e}=\sqrt{2} v_{0} \)
When satellite is in orbit \( K=\frac{1}{2} m v_{0}{ }^{2} \)
For satellite to escape its velocity should be \( v_{e} \)
\( \therefore K E \) To escape, \( K_{e}=\frac{1}{2} m v_{e}^{2}=\frac{1}{2} m\left(\sqrt{2} v_{0}\right)^{2}=2 K \)
\( \therefore \) Extra \( K . E \). required \( =2 K-K=K \)
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