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Chapter 6: Problem 684

If \(R\) is the radius of the earth and \(g\) the acceleration due to gravity on the earth's surface, the mean density of the earth is \(=\ldots \ldots \ldots\) (A) \([(4 \pi \mathrm{G}) /(3 \mathrm{~g} \mathrm{R})]\) (B) \([(3 \pi R) /(4 \mathrm{gG})]\) (C) \([(3 \mathrm{~g}) /(4 \pi \mathrm{RG})]\) (D) \([(\pi R G) /(12 g)]\)

Short Answer

Expert verified
The short answer is: The mean density of the Earth is given by the formula \(\rho = \frac{3g}{4 \pi RG}\).

Step by step solution

01

Gravitational Force Formula

Recall Newton's law of gravitation, which states that the gravitational force F between two objects with masses m1 and m2, separated by a distance r, is given by the equation: \[F = G\frac{m1 \cdot m2}{r^2}\] Here, G is the universal gravitational constant.
02

Define Variables and Find Mass of Earth

Let m be the mass of an object on the surface of the Earth and M be the mass of the Earth. According to Newton's law of gravitation, the gravitational force acting on the object at the surface of Earth is: \[F = G\frac{m \cdot M}{R^2}\] We know that F = m * g (gravitational force is equal to the mass of the object times the acceleration due to gravity) So, \(G\frac{mM}{R^2} = mg\) Now, we can solve for the mass of the Earth (M): \[M = \frac{g \cdot R^2}{G}\]
03

Find Mean Density of the Earth

The mean density (ρ) of the Earth can be calculated using the formula for the mass and volume of a sphere: \[ρ = \frac{M}{(4/3) \pi R^3}\] Substitute the value of M from Step 2 into the mean density formula: \[ρ = \frac{\frac{g \cdot R^2}{G}}{(4/3) \pi R^3}\] Now, simplify the formula: \[ρ = \frac{3g}{4 \pi RG}\] So, the correct answer is: (C) \([(3 \mathrm{~g}) /(4 \pi \mathrm{RG})]\)

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Most popular questions from this chapter

If the radius of earth is \(\mathrm{R}\) then height \({ }^{\prime} \mathrm{h}\) ' at which value of ' \(\mathrm{g}\) ' becomes one-fourth is (A) \(\mathrm{R} / 4\) (B) \(3 \mathrm{R} / 4\) (C) \(\mathrm{R}\) (D) \(\mathrm{R} / 8\)

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite

If \(\mathrm{r}\) denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to (A) \(1^{3 / 2}\) (B) \(\mathrm{r}\) (C) \(r^{1 / 2}\) (D) \(r^{2}\)

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of $2 \mathrm{R}\( from the surface of earth is \)\ldots \ldots \ldots .$ hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

A planet moving along an elliptical orbit is closest to the sun at a distance \(\mathrm{r}_{1}\) and farthest away at a distance of \(\mathrm{r}_{2}\). If \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) are the liner velocities at these points respectively, then the ratio \(\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) is \(\ldots \ldots \ldots \ldots\) (A) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (C) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (D) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)^{2}\)
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