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. In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V
  2. In a tangent galvanometer, a current of 0.1 A produces a deflection of 30⁰.
  3. The surface tension of soap solution is 0.03 N/m. The work done in blowing to form a soap
  4. The ratio of the acceleration for a solid sphere(mass m and radius R)rolling down
  5. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits