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

admin:

Related Questions

  1. Which pair is isotonic?
  2. Four identical rods are joined to from a square. If the temperature difference
  3. In a franhofer diffraction experiment at a single slit using a light of wavelength
  4. The damping force on an oscillator is directly proportional to the velocity.
  5. Photoelectric emision occurs only when the incident light has more than a certain minimum