Question

Assuming earth to be a uniform sphere of radius 6400 km and density 5.5 g/cc. find the value of g on its surface.

Answer 1

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Answer 2

Answer:

The value of g on its surface is $9.8405 m/s^2$.

Explanation:

Gravitational force between the object with mass m and the Earth = F

Mass of earth = M

Radius of the earth ,r= 6400 km = 6,400,000 m

Volume of earth =V = $\frac{4}{3}\pi r^3$

Density of  Earth,D = $\frac{M}{V}$

$D=\frac{M}{\frac{4}{3}\pi r^3}=\frac{3M}{4\pi r^3}$

$\frac{M}{r^3}=\frac{4\pi }{3}\times D$

Density of the earth = D =$5.5 g/cm^3$

$=\frac{5.5\times 0.001 kg}{10^{-6} m^3}=5500 kg/m^3$

$D=5500 kg/m^3$

Gravitational constant = G = $6.674\times 10^{-11} m^3/kg s^2$

F = $G\frac{Mm}{r^2}$

Force on object of mas m on the surface:

$W=mg$

W = F

$mg=G\frac{Mm}{r^2}$

$g=G\frac{M}{r^2}$

$g=G\frac{M}{r^2}\times \frac{r}{r}$

$g=rG\frac{M}{r^3}=rG\times \frac{4\pi }{3}\times D$

$g=6,400,000 m\times 6.674\times 10^{-11} m^3/kg s^2\times \frac{4\times 3.14 }{3}\times 5500 kg/m^3$

$g=9.8405 m/s^2$

The value of g on its surface is $9.8405 m/s^2$.