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NEET PHYSICSGRAVITATIONMedium

Question

The time period of a geostationary satellite is 24 hr24 \text{ hr} at a height 6RE6R_E (RER_E is the radius of the Earth) from the surface of the earth. The time period of another satellite whose height is 2.5RE2.5R_E from the surface will be:

A

62 hr6\sqrt{2} \text{ hr}

B

122 hr12\sqrt{2} \text{ hr}

C

242.5 hr\frac{24}{2.5} \text{ hr}

D

122.5 hr\frac{12}{2.5} \text{ hr}

Step-by-Step Solution

  1. Kepler's Third Law: The square of the time period of revolution of a planet (or satellite) is proportional to the cube of the semi-major axis (radius for circular orbit) of its orbit. [Eq. 7.7, Eq. 7.38] T2r3    T12T22=r13r23T^2 \propto r^3 \implies \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}
  2. Determine Orbital Radii: The distance rr is measured from the center of the Earth. Given RER_E is the Earth's radius.
  • Satellite 1 (Geostationary): Height h1=6REh_1 = 6R_E. Orbit radius r1=RE+h1=RE+6RE=7REr_1 = R_E + h_1 = R_E + 6R_E = 7R_E.
  • Satellite 2: Height h2=2.5REh_2 = 2.5R_E. Orbit radius r2=RE+h2=RE+2.5RE=3.5REr_2 = R_E + h_2 = R_E + 2.5R_E = 3.5R_E.
  1. Calculate Ratio: T2T1=(r2r1)3/2=(3.5RE7RE)3/2=(12)3/2\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2} = \left( \frac{3.5R_E}{7R_E} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} T224=122\frac{T_2}{24} = \frac{1}{2\sqrt{2}}
  2. Solve for T2T_2: T2=2422=122=62 hrT_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ hr}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONperiodgeostationarysatelliteheightradius

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