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NEET PHYSICSGRAVITATIONEasy

Question

A spherical planet has a mass MpM_p and diameter DpD_p. A particle of mass mm falling freely near the surface of this planet will experience acceleration due to gravity equal to:

A

4GMpmDp2\frac{4GM_pm}{D_p^2}

B

4GMpDp2\frac{4GM_p}{D_p^2}

C

GMpmDp2\frac{GM_pm}{D_p^2}

D

GMpDp2\frac{GM_p}{D_p^2}

Step-by-Step Solution

  1. Formula for Acceleration Due to Gravity: The acceleration due to gravity (gg) on the surface of a planet with mass MpM_p and radius RpR_p is derived from Newton's Law of Universal Gravitation. The gravitational force on a mass mm is F=GMpmRp2F = \frac{GM_p m}{R_p^2}. By Newton's second law, force F=mgF = mg. Therefore, the acceleration is g=Fm=GMpRp2g = \frac{F}{m} = \frac{GM_p}{R_p^2} [Equation 7.9].
  2. Substitute Diameter: The problem provides the diameter DpD_p. The radius is half the diameter: Rp=Dp2R_p = \frac{D_p}{2}.
  3. Calculation: Substitute RpR_p into the equation for gg: g=GMp(Dp/2)2=GMpDp2/4=4GMpDp2g = \frac{GM_p}{(D_p/2)^2} = \frac{GM_p}{D_p^2 / 4} = \frac{4GM_p}{D_p^2} Note that acceleration due to gravity is independent of the mass of the falling particle (mm).

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.

PHYSICSGRAVITATIONsphericalplanetdiameterparticlefalling

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