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Chapter 8: Problem 8

Column-I gives the positions of point, column-II ocives expressions for gravitational intensity due to carth. Column-I (Position) (a) At centre of carth (b) At \(rR\) (outsidearth) Column-II (gravitational intensity) (p) \(\frac{G M}{R^{2}}\) (q) 0 (r) \(\frac{G M r}{R^{3}}\) (s) \(\frac{G M}{r^{2}}\).

Short Answer

Expert verified
(a) matches (q), (b) matches (r), (c) matches (p), (d) matches (s).

Step by step solution

01

Understanding Column-I

In Column-I, different positions related to the Earth are given: (a) at the center of the Earth, (b) inside the Earth where radius \( r < R \), (c) at the Earth's surface where \( r = R \), and (d) outside the Earth where \( r > R \).
02

Understanding Column-II

Column-II lists different expressions related to gravitational intensity: (p) \( \frac{G M}{R^{2}} \), (q) 0, (r) \( \frac{G M r}{R^{3}} \), and (s) \( \frac{G M}{r^{2}} \). These expressions relate to the gravitational field at different positions relative to Earth.
03

Match (a) with Column-II

At the center of the Earth (a), the gravitational intensity becomes zero because the gravitational forces from all sides cancel each other out. Thus, match (a) with (q) 0.
04

Match (b) with Column-II

Inside the Earth at a distance \( r < R \), the gravitational intensity is proportional to \( r \) and given by \( \frac{G M r}{R^{3}} \) due to the shell theorem. Thus, match (b) with (r) \( \frac{G M r}{R^{3}} \).
05

Match (c) with Column-II

At the surface of the Earth \( (r = R) \), the gravitational intensity is \( \frac{G M}{R^{2}} \) because it is determined by the entire mass of the Earth acting at its radius. Match (c) with (p) \( \frac{G M}{R^{2}} \).
06

Match (d) with Column-II

Outside the Earth where \( r > R \), the gravitational intensity decreases with the square of the distance and is given by \( \frac{G M}{r^{2}} \) due to the inverse square law. Match (d) with (s) \( \frac{G M}{r^{2}} \).

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Key Concepts

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

Shell Theorem
The shell theorem is an important concept in gravity and physics. This theorem simplifies how we understand gravitational forces inside and outside a spherical shell of mass.
  • Inside a uniform spherical shell of mass, the gravitational force on a point mass inside the shell is zero.
  • Outside the shell, the gravitational force behaves as if the entire mass is concentrated at the center of the shell.
For example, when considering the Earth, if you're inside the Earth's radius, some parts of the Earth pull you one way, while others pull you in the opposite direction. Thanks to the shell theorem, these forces cancel each other out at the center, which is why the gravitational intensity is zero at Earth's core.
This theorem saves us a lot of work in calculations, especially when defining gravitational intensity at various positions within or around a massive sphere like the Earth.
Gravitational Field
The gravitational field is a vector field describing the gravitational force exerted by a mass on objects in the surrounding space. This field is crucial for understanding how objects interact with one another through gravity.
Gravitational field intensity at a point is defined as the gravitational force experienced by a unit mass placed at that point.
  • At the center of the Earth, the gravitational field intensity is zero, as the forces from all sides cancel out.
  • Inside the Earth but away from the center, the field intensity increases proportionally with distance from the center.
  • At the surface and outside the Earth, it follows the inverse square law, which we'll explore next.
Understanding gravitational fields helps us predict how an object will move under gravitational influence. Every point in space in the vicinity of a massive object like Earth has a certain field intensity, which dictates the gravitational 'pull' exerted there.
Inverse Square Law
The inverse square law is a fundamental principle that applies to various forces in nature, including gravity. This law states that the intensity of a force (such as gravitational force) diminishes proportionally to the square of the distance from the source.
In the context of gravity, this means:
  • As you move farther from the source of gravity (like Earth), the gravitational intensity decreases sharply.
  • When at a distance greater than Earth's radius, the gravitational force is calculated using \( \frac{G M}{r^2} \).
The inverse square law explains why objects feel less gravitational pull when they are far away from a massive object. This behavior is observable in many areas of physics, confirming that fundamental forces weaken rapidly with increasing distance. It's a vital concept when calculating gravitational forces for objects outside Earth鈥檚 immediate vicinity.
Earth's Gravitational Field
Earth鈥檚 gravitational field is the region around Earth where objects experience a gravitational pull towards it. Understanding its behavior at different positions is key for tasks like space missions, satellite operations, and everyday physics problems.
  • At Earth's center, the gravitational field is zero due to symmetric forces cancelling out.
  • Within the Earth, from the center upwards towards the surface, the gravitational field increases linearly.
  • At the Earth's surface, this field is what causes objects to have weight, calculated using \( \frac{G M}{R^2} \).
  • Beyond Earth's surface, the gravitational field weakens following the inverse square law.
Earth鈥檚 gravitational field not only affects terrestrial objects but also plays a significant role in outer space activities. Its predictable nature allows scientists and engineers to design structures and missions that take into account how gravity will influence their movement and trajectory.

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Most popular questions from this chapter

Three particles \(A, B\) and \(C\) each of mass \(m\) lie on vertices of an equilateral triangle of side \(a\). The system is released from rest. What will be the net momentum of two particles \(A\) and \(B\), when separation between the particles remains \(a / 2 ?\) (a) \(\sqrt{2 G m^{3} / a}\) (b) \(\sqrt{G m^{3} / a}\) (c) \(\sqrt{6 G m^{3} / a}\) (d) \(\sqrt{2 G m^{3} / 3 a}\)

\(R\) is the radius of carth and \(\omega\) is ils angular velocity and \(g_{t}\) is the value of accelcration duc to gravity at cquator. The effoctive value of acccleration due to gravity at the latitude \(\lambda-30^{\circ}\) will be cqual to (a) \(g_{E}-\frac{3}{4} \omega^{2} R\) (b) \(g_{E}+\frac{3}{4} R \omega^{2}\) (c) \(g_{r}+\frac{\omega^{2} R}{4}\) (d) \(g_{E}-\frac{\omega^{2} R}{4}\)

A solid sphere of uniform density and radius 4 units is located with its centre at the origin \(O\) of coordinates. Two spheres of equal radii 1 unit, with their centres at \(A(-2,0,0)\) and \(B(2,0,0)\) respectively, are taken out of the solid leaving behind spherical cavitics as shown in the figurc. Then, (a) the gravitational force due to this object at the origin is zero. (b) the gravitational force at the point \(B(2,0,0)\) is zero. (c) the gravitational potential is the same at all points of the circle \(z^{2}+y^{2}-36\) (d) the gravitational potential is same at all points of the circle \(y^{2}+z^{2}-4\)

Which of the following statements is/are true for a stationary satellite of the carth? (a) I is stationary in spaco (b) Its time period is 24 hours (c) Its angular speed is cqual to that of carth about its own axis (d) It rovolves from west to cast

The magnitudes of the gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a unilorm sphere of radius \(R\) and mass \(M\) arc \(F_{1}\) and \(F_{2}\) respectively. Then (a) \(\frac{F_{1}}{F_{2}}-\frac{r_{1}}{r_{2}}\) if \(r_{1}R\) and \(r_{2}>R\) (c) \(\frac{F_{1}}{F_{2}}-\frac{r_{1}^{3}}{r_{2}^{3}}\) if \(r_{1}
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