Gravitation - Result Question 41

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44. A geostationary satellite is orbiting the earth at a height of $5 R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2 R$ from the surface of the earth is :

======= ####44. A geostationary satellite is orbiting the earth at a height of $5 R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2 R$ from the surface of the earth is :

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/gravitation/gravitation—result-question-41.md (a) 5

(b) 10

(c) $6 \sqrt{2}$

(d) $\frac{6}{\sqrt{2}}$

[2012]

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Answer:

Correct Answer: 44. (c)

Solution:

  1. (c) According to Kelpner’s law of period $T^{2} \propto R^{3}$

$\frac{T_1^{2}}{T_2^{2}}=\frac{R_1^{3}}{R_2^{3}}=\frac{(6 R)^{3}}{(3 R)^{3}}=8$

$\frac{24 \times 24}{T_2^{2}}=8$

$T_2^{2}=\frac{24 \times 24}{8}=72=36 \times 2$

$T_2=6 \sqrt{2}$