Navigate

The acceleration due to gravity near the surface of a planet of radius R and density d

The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to

Options

(a) d/R²
(b) dR²
(c) dR
(d) d/R

Correct Answer:

dR

Explanation:

g=GM/R²
(M=Mass of the earth); (R=Distance of body from centre of earth)
g=G Volume x density / R²
Volume of the sphere=4/3 πR³
Therefore, g=G.4/3 πR³.d / R²
g=G 4/3 πRd
g=4πG/3.dR
g is proportional to dR

MP2002
admin:

Related Questions

  1. The photoelectric threshold wevelength for potassium (work function being 2 eV) is
  2. A solenoid of length l metre has self inductance L henry. If number of turns
  3. In a vessel, the gas is at a pressure P. If the mass of all the molecules
  4. If the distance between the plates of two capacitor is increased then capacitance
  5. Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line