MCQ
The kinetic energy of the satellite in a circular orbit with speed $v$ is given as
- ✓$KE =\frac{G m M_e}{2\left(R_e+h\right)}$
- B$KE =-\frac{1}{2} m v^2$
- C$KE =\frac{G m M_e}{\left(R_e+h\right)}$
- D$KE =\frac{-G m M_e}{2\left(R_e+h\right)}$
✓
Answer
Correct option: A.
$KE =\frac{G m M_e}{2\left(R_e+h\right)}$
$\text { Explanation: } KE \text { of satellite }=\frac{1}{2} mv ^2$
$=\frac{1}{2} m\left(\sqrt{\frac{G M_e}{\left(R_e+h\right)}}\right)^2$
$=\frac{1}{2} \frac{G m M_e}{\left(R_e+h\right)}$
$=\frac{1}{2} m\left(\sqrt{\frac{G M_e}{\left(R_e+h\right)}}\right)^2$
$=\frac{1}{2} \frac{G m M_e}{\left(R_e+h\right)}$
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