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At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)

Short Answer

Expert verified
The distance at which the acceleration due to gravity is half that of its value at the Earth's surface is \(r = 1.414 R\), which corresponds to option (C).

Step by step solution

01

Understand the gravity formula

The acceleration due to gravity, g, at a distance r from the center of the Earth is given by the formula: \[g(r) = \frac{GM}{r^2}\] where G is the gravitational constant and M is the mass of the Earth.
02

Know the gravity on the surface of the Earth

We're given that at the surface of the Earth, the value of acceleration due to gravity is \(g_s\). Using the formula, we can write it as: \[g_s = \frac{GM}{R^2}\]
03

Set up the equation for half the gravity value

We are asked to find the distance from the center of the Earth where the acceleration due to gravity is half that of the surface. So at that distance, let's call it r, the gravity will be equal to \(\frac{1}{2}g_s\). Therefore, using the formula, we can write it as: \[\frac{1}{2}g_s = \frac{GM}{r^2}\]
04

Solve for the required distance r

Now we have the equations for the gravity at the surface of the Earth and for it being half at the distance r. We can solve for r using these. Divide the equation for half the gravity value by the equation for the gravity value at the surface: \[\frac{\frac{1}{2}g_s}{g_s} = \frac{\frac{GM}{r^2}}{\frac{GM}{R^2}}\] The gravitational constant G and Earth's mass M cancel out within the fractions, leading to: \[\frac{\frac{1}{2}}{1} = \frac{R^2}{r^2}\] Next, we get the following equation: \[\frac{1}{2} = \frac{R^2}{r^2}\] Cross multiply and solve for r: \[r^2 = 2R^2\] \[r = \sqrt{2R^2} = R\sqrt{2}\] So the distance at which gravity is half its value at the surface of the Earth is \(r = 1.414 R\), which corresponds to option (C).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational constant
The gravitational constant, denoted by the symbol G, is a fundamental constant of nature crucial to our understanding of gravitational forces. It appears in the universal law of gravitation formulated by Isaac Newton. This constant provides the proportionality factor that allows us to calculate the gravitational force exerted between two masses. The value of G is approximately \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\).

It might seem abstract, but G ensures that we can quantitatively describe how objects with mass attract each other. It's important to understand that G is not dependent on the type or size of the masses involved or the distance between them. Its universality is the reason why it is called a constant.
  • Utilized in Newton's law of universal gravitation.
  • Essential for determining forces between celestial bodies.
Studying the gravitational constant helps explain the dynamics of planets and stars. It plays a key role in the calculations that predict the positions and movements of celestial objects, essential for space exploration.
Radius of Earth
The radius of Earth is vital for calculating gravitational forces, understanding the planet’s physical dimensions, and more. Generally approximated as \(6,371\, \text{km}\), it represents an average value between the radius at the poles and at the equator.

When considering gravitational calculations, the radius of the Earth, often denoted as R, becomes a reference point for understanding how gravity decreases with distance from the center of the Earth. As you move away from the Earth's surface, the gravity decreases due to the increase in distance. Thus, R is pivotal for deriving expressions related to varying gravitational acceleration.
  • Central to calculations of gravitational force.
  • Assists in modeling Earth's gravitational field.
Knowing Earth's radius helps quantify how other forces, like air resistance, act on objects. Additionally, it serves as a stepping-stone for understanding more complex planetary metrics and behaviors.
Newton's law of universal gravitation
Newton's law of universal gravitation is a cornerstone in classical physics, describing the gravitational attraction between masses. This law asserts that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of the force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Mathematically, it is represented as:\[ F = \frac{GMm}{r^2} \] where:
  • \(F\) is the gravitational force between two masses.
  • \(M\) and \(m\) are the two masses.
  • \(r\) is the distance between the centers of the two masses.
This law is fundamental in explaining not just how objects fall to Earth, but also how celestial bodies like planets and moons interact with each other in space.

Newton's insight dramatically altered humanity's understanding of the universe, offering a unified framework where the same principles of motion dictate phenomena both on Earth and in the cosmos. This framework is essential for fields ranging from engineering to space travel.

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Most popular questions from this chapter

Gravitational potential at any point inside a spherical shall is uniform and is given by \(-(\mathrm{GM} / \mathrm{R})\) where \(\mathrm{M}\) is the mass of shell and \(\mathrm{R}\) its radius. At the center solid sphere, potential is \([-\\{(3 \mathrm{GM}) /(2 \mathrm{R})\\}]\)(1) There is a concentric hole of radius \(\mathrm{R}\) in a solid sphere of radius \(2 \mathrm{R}\) mass of the remaining portion is \(\mathrm{M}\) what is the gravitational at center? (A) \(-[(3 \mathrm{GM}) / 7 \mathrm{R}]\) (B) \(-[(5 \mathrm{GM}) / 7 \mathrm{R}]\) (C) \(-[(7 \mathrm{GM}) / 14]\) (D) \(-[(9 \mathrm{GM}) /(14 \mathrm{R})]\)

Two satellites of mass \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\left(\mathrm{~m}_{1}=\mathrm{m}_{2}\right)\) are revolving round the earth in circular orbits of \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{1}>\mathrm{r}_{2}\right)\) respectively. Which of the following statement is true regarding their speeds \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) (A) \(\mathrm{V}_{1}=\mathrm{V}_{2}\) (B) \(\mathrm{V}_{1}<\mathrm{V}_{2}\) (C) \(\mathrm{V}_{1}>\mathrm{V}_{2}\) (D) \(\left(\mathrm{V}_{1} / \mathrm{r}_{1}\right)=\left(\mathrm{V}_{2} / \mathrm{r}_{2}\right)\)

The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is (A) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\) (B) \(3 \mathrm{~V}_{\mathrm{e}}\) (C) \(\sqrt{2} \mathrm{~V}_{\mathrm{e}}\) (D) \(2 \mathrm{~V}_{\mathrm{e}}\)

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) \(12 \mathrm{v}\)

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)

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