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Chapter 6: Problem 708

Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be \(-m s^{-2}\) \(\left(\mathrm{g}=9.8 \mathrm{~ms}^{2}\right)\) (A) \(19.6\) (B) \(9.8\) (C) \(4.9\) (D) \(2.45\)

Short Answer

Expert verified
The new acceleration due to gravity is \(19.6 \text{ m/s²}\), which corresponds to the choice (A).

Step by step solution

01

Write the formula for acceleration due to gravity

The formula for acceleration due to gravity is: \[g = \frac{GM}{r^2}\] Where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the Earth.
02

Write the mass-density relationship formula

The mass-density relationship is given by: \[M = \rho V\] Where M is the mass, ρ is the density, and V is the volume.
03

Substitute the mass-density relationship in the gravity formula

Replace M in the gravity formula using the mass-density relationship: \[g = \frac{G \cdot (\rho V)}{r^2}\]
04

Rewrite the formula in terms of radius

Since the radius of the Earth is constant, we can rewrite the volume in terms of radius (using the formula for the volume of a sphere, \(V = \frac{4}{3} \pi r^3\)): \[g = \frac{G \cdot (\rho \cdot \frac{4}{3} \pi r^3)}{r^2} =\frac{4}{3}G\pi r \rho\]
05

Calculate acceleration due to gravity when Earth's density is doubled

Now, we double the density (2ρ) and find the new acceleration due to gravity: \[g' = \frac{4}{3}G\pi r (2 \rho) \]
06

Calculate the ratio of the new gravity to the initial gravity

To find the new gravity, observe the ratio between the new gravity and the initial gravity: \[\frac{g'}{g} = \frac{\frac{4}{3}G\pi r (2 \rho)}{\frac{4}{3}G\pi r \rho} = 2\]
07

Calculate the new acceleration due to gravity

Since the ratio of the new gravity to the initial gravity is 2, the new acceleration due to gravity (g') is twice the initial gravity (9.8 m/s²): \[g' = 2 \cdot 9.8 = 19.6 \text{ m/s²}\] Answer: The new acceleration due to gravity is 19.6 m/s², which corresponds to the choice (A).

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Most popular questions from this chapter

If the density of small planet is that of the same as that of the earth while the radius of the planet is \(0.2\) times that of the earth, the gravitational acceleration on the surface of the planet is (A) \(0.2 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(2 \mathrm{~g}\) (D) \(4 \mathrm{~g}\)

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If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ' \(\mathrm{g}\) ' on the earth is \(9.8 \mathrm{~ms}^{-2}\), then the value of \(\mathrm{g}^{\prime}\) on that planet is $=\ldots \ldots \ldots \mathrm{ms}^{-2}$ (A) \(4.9\) (B) \(0.98\) (C) \(0.49\) (D) 49

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