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Chapter 6: Problem 715

The acceleration of a body due to the attraction of the earth (radius R) at a distance \(2 \mathrm{R}\) from the surface of the earth is \(=\) (g \(=\overline{\text { acceleration due to gravity at the surface of earth })}\) (A) \(\mathrm{g} / 9\) (B) \(\mathrm{g} / 3\) (C) \(\mathrm{g} / 4\) (D) 9

Short Answer

Expert verified
The acceleration of a body due to the attraction of the earth at a distance \(2R\) from the surface of the earth is (A) \(g/9\).

Step by step solution

01

Recall gravitational force formula

The formula for gravitational force between two masses (m1 and m2) separated by distance r is given by: \(F = G\frac{m1*m2}{r^2}\) where G is the gravitational constant. In this case, m1 is the mass of the Earth (M) and m2 is the mass of the body (m). The distance r in our case is the sum of Earth's radius and the given distance from the surface, so r = R + 2R = 3R.
02

Find gravitational acceleration at 2R distance

Divide the gravitational force by the mass of the body (m) to get the acceleration due to gravity (a) at distance 2R from the surface of the Earth: \(a = \frac{F}{m} = \frac{G\frac{M*m}{(3R)^2}}{m} = G\frac{M}{9R^2}\)
03

Relate the gravitational acceleration at 2R to the Earth's surface

We know that the acceleration due to gravity at the surface of the Earth (g) is given by: \(g = G\frac{M}{R^2}\) Now, we'll divide the acceleration at 2R (a) by the acceleration due to gravity at the surface of the Earth (g): \(\frac{a}{g} = \frac{G\frac{M}{9R^2}}{G\frac{M}{R^2}} = \frac{1}{9}\)
04

Choose the correct answer

Since the acceleration at 2R distance from the surface of the Earth is g/9, the correct answer is: (A) g/9

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Most popular questions from this chapter

The period of a satellite in circular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C)

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