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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

Potential energy of a satellite having mass \(\mathrm{m}\) and rotating at a height of \(6.4 \times 10^{6} \mathrm{~m}\) from the surface of earth (A) \(-0.5 \mathrm{mg} \operatorname{Re}\) (B) \(-\mathrm{mg} \mathrm{Re}\) (C) \(-2 \mathrm{mgRe}\) (D) \(4 \mathrm{mgRe}\)

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

If the value of ' \(\mathrm{g}\) ' acceleration due to gravity, at earth surface is \(10 \mathrm{~ms}^{-2}\). its value in \(\mathrm{ms}^{-2}\) at the center of earth, which is assumed to be a sphere of Radius ' \(\mathrm{R}\) 'meter and uniform density is (A) 5 (B) \(10 / \mathrm{R}\) (C) \(10 / 2 \mathrm{R}\) (D) zero

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The value of acc. due to gravity \((\mathrm{g})\) does not depend upon mass of the body Reason: This follows from \(\mathrm{g}=\left[(\mathrm{GM}) / \mathrm{R}^{2}\right]\), where \(\mathrm{M}\) is mass of planet (earth) and \(\mathrm{R}\) is radius of planet (earth) (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively If the speed of satellite \(\mathrm{A}\) is \(3 \mathrm{v}\), then speed of satellite \(\mathrm{B}\) is (A) \((3 \mathrm{v} / 2)\) (B) \((4 \mathrm{v} / 2)\) (C) \(6 \mathrm{v}\) (D) \(12 \mathrm{v}\)
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