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NEET 2012
Exam
NEET
Year
2012
Subject
Physics
Chapter
Gravitation

NEET 2012 · Physics · Gravitation

A geostationary satellite is orbiting the earth at a height of 5 R...

A geostationary satellite is orbiting the earth at a height of 5R5 R above that surface of the earth, RR being the radius of the earth. The time period of another satellite in hours at a height of 2R2 R from the surface of the earth is

Options

  1. A5
  2. B10
  3. C626 \sqrt{2}
  4. D6/26 / \sqrt{2}

Correct answer

C. 626 \sqrt{2}

Step-by-step solution

From Keplar third's law T2∝r3T^2 \propto r^3 Hence, T12∝r13T_1^2 \propto r_1^3 So, T22∝r23T22T12=r23r13=(3R)3(6R)3\begin{aligned} & T_2^2 \propto r_2^3 \\ & \frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3} \\ &=\frac{(3 R)^3}{(6 R)^3} \end{aligned}or T22T12=18T22=18T12T2=2422=62h\begin{aligned} & \text{or } \frac{T_2^2}{T_1^2}=\frac{1}{8} \\ & T_2^2=\frac{1}{8} T_1^2 \\ & T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{ h} \end{aligned}$
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