The acceleration due to gravity at the surface of the earth (mass M and radius R) is proportional to __________.

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RRB ALP CBT I 17 Aug 2018 Shift 3 Official Paper
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  1. \(\frac{M}{R}\)
  2. MR
  3. \(\frac{M}{R^2}\)
  4. \(\frac{M^2}{R}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{M}{R^2}\)
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Detailed Solution

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The correct answer is M/R2.

Key Points

  • The force acting on an object due to gravity is given by f = mg.
  • Where f is the force acting on the object, g is the acceleration due to gravity, and m is the mass of the object.
  • According to the law of universal gravitation,
  • f = GmM/(r+h)2.
  • Where,
    f = force between two bodies,
    G = universal gravitation constant (6.67×10-11 Nm2/kg2)
    m = mass of an object,
    M = mass of earth,
    r = radius of the Earth.
    h = height the object is from the earth's surface.
  • Since the height (h) is negligible relative to the radius of the earth, we repeat the equation as follows,
    f = GmM/r2
  • Now balance the two expressions,
    mg = GmM/r2
    g = GM /r2
  • Therefore, the formula for the acceleration due to gravity is given by, g = GM/r2
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