Question 303: Gravitation

Question

Considering earth to be a sphere of radius '$$R$$' having uniform density '$$\rho$$', then value of acceleration due to gravity '$$g$$' in terms of $$R, \rho$$ and $$\mathrm{G}$$ is

Accepted Answer

To derive an expression for the acceleration due to gravity 'g' at the surface of the Earth, in terms of its radius 'R', its uniform density 'ÃÂ', and the gravitational constant 'G', we can start by calculating the mass 'M' of the Earth in terms of its density and volume.
The volume 'V' of a sphere is given by
$$ V = \frac{4}{3} \pi R^3 $$
For a sphere with uniform density 'ÃÂ', the mass 'M' can be calculated by multiplying the volume by the density:
$$ M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right) $$
Now that we have the mass, we can use Newton's law of universal gravitation to find the force 'F' exerted on a mass 'm' at the surface of the Earth:
$$ F = \frac{G M m}{R^2} $$
The acceleration 'g' due to gravity at the Earth's surface is simply the force per unit mass 'm':
$$ g = \frac{F}{m} = \frac{G M}{R^2} $$
Substituting the expression for 'M' in terms of 'ÃÂ' and 'R', we get:
$$ g = \frac{G \left( \rho \left( \frac{4}{3} \pi R^3 \right) \right)}{R^2} $$
When we simplify this expression, we have:
$$ g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} $$
$$ g = \frac{4}{3} \pi G \rho R $$
Therefore, the correct answer is:
Option C
$$ g = \frac{4}{3} \pi \rho GR $$

Answer

$$g=\sqrt{\frac{3 \pi R}{\rho G}}$$

Answer

$$\mathrm{g}=\sqrt{\frac{4}{3} \pi \rho \mathrm{GR}}$$

Answer

$$\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}$$

Answer

$$g=\frac{G M}{\rho R^2}$$

Gravitation

Considering earth to be a sphere of radius '$R' having uniform density '\rho', then value of acceleration due to gravity 'g' in terms of R, \rho and \mathrm{G}$ is

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