Gravitation Ques 2
2. The time period of a geostationary satellite is $24 h$, at a height $6 R_E\left(R_E\right.$ is radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathrm{R}_{\mathrm{E}}$ from surface will be,
[NEET Odisha 2019]
(a) $\frac{12}{2.5}$ $ h$
(b) $6 \sqrt{2}$ $ h$
(c) $12 \sqrt{2}$ $ h$
(d) $\frac{24}{2.5}$ $ h$
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Answer:
Correct Answer: 2.(b)
Solution: (b) $T^2 \propto r^3$
$T^2 \propto\left(R_E+h\right)^3$
$\frac{T_1^2}{T_2^2} =\frac{\left(R_E+6 R_E\right)^3}{\left(R_E+2.5 R_E\right)^3}$
$\frac{T_1^2}{T_2^2} =\frac{7^3}{\left(\frac{7}{2}\right)^3}$
$T_2 =6 \sqrt{2}$ $ h$
BETA
