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Chapter 6: Problem 700

A body weight \(500 \mathrm{~N}\) on the surface of the earth. How much would it weight half way below the surface of earth (A) \(125 \mathrm{~N}\) (B) \(250 \mathrm{~N}\) (C) \(500 \mathrm{~N}\) (D) \(1000 \mathrm{~N}\)

Short Answer

Expert verified
The weight of the body half-way below the surface of the Earth is approximately 250 N. The correct answer is (B) \(250 \mathrm{~N}\).

Step by step solution

01

Write down the gravitational force formula

The formula for the gravitational force between two objects is given by: \[F = G \frac{m_1 m_2}{r^2}\] Where: - F is the gravitational force - G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}\)) - \(m_1\) and \(m_2\) are the masses of the two objects - r is the distance between the centers of the objects In this exercise, we will consider Earth as one object and the body with a weight of 500 N as the other object.
02

Determine the mass of the body

First, we need to find the mass of the body. To do this, we can use the formula for weight: \[W = mg\] Where: - W is the weight of the body (500 N) - m is the mass of the body (which we want to find) - g is the acceleration due to gravity on Earth's surface (\(9.81\mathrm{m/s^2}\)) Solving for m: \[m = \frac{W}{g}\]
03

Calculate the mass of the body

Using the values given, we can now calculate the mass of the body: \[m = \frac{500\mathrm{~N}}{9.81\mathrm{m/s^2}} \approx 51\mathrm{kg}\]
04

Find the new distance to the center of the Earth

Now, we need to find the new distance to the center of the Earth when the body is half-way below the surface. Let's call this new distance \(r'\). The radius of the Earth is approximately 6371 km. Therefore, the new distance to the center of the Earth is half of this value: \[r' = \frac{1}{2} (6371\mathrm{km}) = 3185.5\mathrm{km}\] Make sure to convert this value to meters: \[r' = 3185.5\mathrm{km} \times 1000\mathrm{m/km} = 3185500\mathrm{m}\]
05

Calculate the new gravitational force

Using the gravitational force formula, we can now calculate the new gravitational force acting on the body at half-way below the surface: \[F' = G \frac{m_1 m_2}{{r'}^2}\] Here, \(m_1\) is the body's mass (51 kg) and \(m_2\) is Earth's mass (\(5.972 \times 10^{24}\mathrm{kg}\)). Using the values, we get: \[F' \approx \frac{6.674 \times 10^{-11} \mathrm{Nm^2/kg^2} \times 51\mathrm{kg} \times 5.972 \times 10^{24}\mathrm{kg}}{{(3185500\mathrm{m})}^2}\] \[F' \approx 250\mathrm{~N}\]
06

Choose the correct answer

From our calculation, the weight of the body half-way below the surface of the Earth is approximately 250 N, thus the correct answer is: (B) \(250 \mathrm{~N}\)

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Most popular questions from this chapter

The additional K.E. to be provided to a satellite of mass \(\mathrm{m}\) revolving around a planet of mass \(\mathrm{M}\), to transfer it from a circular orbit of radius \(\mathrm{R}_{1}\) to another radius \(\mathrm{R}_{2}\left(\mathrm{R}_{2}>\mathrm{R}_{1}\right)\) is (A) $\operatorname{GMm}\left[\left(1 / R_{1}^{2}\right)-\left(1 / R_{2}^{2}\right)\right]$ \(\operatorname{GMm}\left[\left(1 / R_{1}\right)-\left(1 / R_{2}\right)\right]\) (C) $2 \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$ (D) $(1 / 2) \mathrm{GMm}\left[\left(1 / \mathrm{R}_{1}\right)-\left(1 / \mathrm{R}_{2}\right)\right]$

The acceleration due to gravity near the surface of a planet of radius \(\mathrm{R}\) and density \(\mathrm{d}\) is proportional to (A) \(\mathrm{d} / \mathrm{R}^{2}\) (B) \(\mathrm{d} \mathrm{R}^{2}\) (C) \(\mathrm{dR}\) (D) \(\mathrm{d} / \mathrm{R}\)

A satellite of mass \(\mathrm{m}\) is orbiting close to the surface of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) has a K.E. \(\mathrm{K}\). The corresponding \(\mathrm{K} . \mathrm{E}\). of satellite to escape from the earth's gravitational field is (A) \(\mathrm{K}\) (B) \(2 \mathrm{~K}\) (C) \(\mathrm{mg} \mathrm{R}\) (D) \(\mathrm{m} \mathrm{K}\)

The moon's radius is \(1 / 4\) that of earth and its mass is \(1 / 80\) times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is (A) \(g / 4\) (B) \(\mathrm{g} / 5\) (c) \(\mathrm{g} / 6\) (D) \(\mathrm{g} / 8\)

Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$
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