The value of g (acceleration due to gravity) at earth's surface is 10 ms-2 its value in ms-2 at centre of earth which is assumed to be a sphere of radius R metre and uniform mass density is
The value of g (acceleration due to gravity) at earth's surface is 10 ms-2 its value in ms-2 at centre of earth which is assumed to be a sphere of radius R metre and uniform mass density is
Dear student,
lets assume that the Earth is a sphere whose density is spherically symmetric.
Now consider a mass m at some radius rrfrom the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on mm of all mass with radii greater than r exert no net force on it. It follows that only the mass with radii less than or equal to r contribute to the gravitational force on m, which, by the Law of Gravitation is
F(r)=G ×M(r)×m/r²
where M(r) is the mass of stuff at radii less than or equal to rr. Notice, then, that F(r)will be an increasing function of rr(and will decrease as r→0), provided M(r)/r² is an increasing function of r.
Now, If the Earth were uniformly dense with density ρ0, then the mass within a radius r would be
M(r)=
4×π×r³×po(density)/3
namely just the density times the volume of a sphere of radius r, and in this case the strength of the gravitational field as a function of radius would be
g(r)=F(r)/m=
G×(1/r²)×(4/3)×π×r³×po
=(4/3)×π×g×po×r
So in this case, it would be true that the strength of the gravitational field would decrease with increasing depth and as it reaches centre of earth r becomes zero, therefore, g also becomes zero (0). So, this is the value of g in the centre of the earth, considering it to be a uniform sphere. However this is not true as earth is not a uniform sphere.
thank you.
lets
assume that the Earth is a sphere whose density is spherically symmetric.
Now consider a mass m at some radius rrfrom the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on mm of all mass with radii greater than r exert no net force on it. It follows that only the mass with radii less than or equal to r contribute to the gravitational force on m, which, by the Law of Gravitation is
F(r)=G ×M(r)×m/r²
where M(r) is the mass of stuff at radii less than or equal to rr. Notice, then, that F(r)will be an increasing function of rr(and will decrease as r→0), provided M(r)/r² is an increasing function of r.
Now, If the Earth were uniformly dense with density ρ0, then the mass within a radius r would be
M(r)=
4×π×r³×po(density)/3
namely just the density times the volume of a sphere of radius r, and in this case the strength of the gravitational field as a function of radius would be
g(r)=F(r)/m=
G×(1/r²)×(4/3)×π×r³×po
=(4/3)×π×g×po×r
So in this case, it would be true that the strength of the gravitational field would decrease with increasing depth and as it reaches centre of earth r becomes zero, therefore, g also becomes zero (0). So, this is the value of g in the centre of the earth, considering it to be a uniform sphere. However this is not true as earth is not a uniform sphere.
thank you.
