Question

Density of earth is 5478 kg/m3.assuming the earth to be homogeneous sphere. find the value Value of ‘g’ on the surface of the earth.(Given: G=6.673x10-11 Nm2/kg2,R=6400km)

Answer 1

Answer:

9.799m/s^2

Explanation:

density=mass/volume

m=d*v=5478*(4/3)*pi(6400000)^3=6.015*10^23

now

g=Gm/r^2

g=6.673*10^-11*6.015*10^23/(6400000)^2=9.799

Answer 2

The value of g is $g=9.8\ m/s^2$.

Explanation:

Given that,

Density of Earth, $d=5478\ kg/m^3$

We know that,

The value of gravitational constant, $G=6.673\times 10^{-11}\ Nm^2/kg^2$

Radius of the Earth, R = 6400 km

Density of an object is given by its mass per unit volume. It is given by :

$d=\dfrac{M}{V}$

$M=d\times \dfrac{4}{3}\pi R^3$

The value of g is given by the below relation as :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{4}{3}\pi dR^3\times \dfrac{GM}{R^2}$

$g=\dfrac{4}{3}\pi dGR$

$g=\dfrac{4}{3}\pi \times 5478 \times 6.67\times 10^{-11} \times 64\times 10^5$

$g=9.795\ m/s^2$

or

$g=9.8\ m/s^2$

So, the value of g is $g=9.8\ m/s^2$. Hence, this is the required solution.

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