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. If |A x B|= √3 A.B , then the value of |A + B| is
  2. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV.
  3. At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
  4. 540 g of ice at 0° C is mixed with 540 g of water at 80° C. The final temperature
  5. In a uniform circular motion,