Question

a spherical hole of radius r/2 is excavated from asteriod of mass m find gravitational accelaration at a point on the surface of asteroid just above the excavation

Answer 1

Answer:

ga = G Ma / R^2        gravitational acceleration due to asteroid alone

gb = G Mb / (R/2)^2   gravitational acceleration due to sphere of radius R/2

g = ga - gb       acceleration with spherical hole in Ma

g = G / R^2 * ( Ma - 4 Mb)     (I)

Va = 4/3 pi R^3     volume of asteroid

Vb = 4/3 pi (R/2)^3 = pi R^3 / 6

Va - Vb = pi R^3 * (4/3 - 1/6) = 7 /6 pi R^3

Ma - Mb = d * 7/6 pi R^3   where d is density

Ma = 4/3 pi R^3 d

d = 3 Ma / (4 pi R^3)

Ma - Mb = 7 /6 pi R^3 * 3 Ma / (4 pi R^3)

Ma - Mb = 21/24 Ma = 7/8 Ma

so Mb = Ma / 8  

(Ma - 4 Mb) = Ma / 2

g = G * (Ma - 4 Mb) / R^2          (I) above

g =  G Ma / (2 R^2)

Answer 2

Answer:GM/2R^2

Explanation:

Ps:don't mind the diagram tho

Hope this helps