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This chapter lays the foundation for understanding the fundamental force that governs the motion of celestial objects and everyday interactions with gravity. We'll explore Newton's law of universal gravitation, delve into the vector nature of gravitational force, and understand the principle of superposition for combining gravitational interactions.
Sir Isaac Newton revolutionized our understanding of gravity with his universal law of gravitation. This law states:
Mathematically, it can be expressed as:
F = G (m1 m2) / r^2
where:
Key Points:
While the scalar form of the law provides the magnitude of the force, the vector form specifies its direction. The gravitational force acts along the line joining the centers of the two masses. We can express it as:
F = G (m1 m2) * ((r1 - r2) / |r1 - r2|^2)
where:
Table 1: Comparison of Scalar and Vector Forms
Feature Scalar Form Vector Form Magnitude F Direction Not specified Along the line joining centers
Understanding the vector nature of gravity becomes crucial when dealing with situations involving multiple interacting masses.
The principle of superposition states that the net gravitational force acting on a particle due to several other particles is the vector sum of the individual gravitational forces exerted by each particle.
In simpler terms, for a particle interacting with multiple other masses, the total force it experiences is the sum of the individual forces exerted by each mass, taking into account their directions.
Mathematically, the net force can be found using vector addition:
F_net = Σ (G (m_i m) / r_i^2) * ((r_i - r) / |r_i - r|^2)
where:
The gravitational field at a point in space is defined as the gravitational force per unit mass experienced by a small test mass placed at that point. It's a vector field, meaning it has both magnitude and direction. The direction of the gravitational field coincides with the direction of the force a test mass would experience.
Key Points:
The gravitational field created by various objects can be calculated using the following formula:
g = G * (M / r^2)
where:
The following sections will explore the gravitational field due to various objects:
A point mass is an object whose entire mass is concentrated at a single point. The gravitational field due to a point mass is:
g = G * (M / r^2)
This is the most basic form of the formula, applicable to any object considered a point mass for a specific calculation.
For a system of point masses, the principle of superposition applies. The net gravitational field at a point is the vector sum of the individual fields created by each point mass.
On the axis of a finite rod:
The gravitational field due to a finite rod on its axis can be calculated using integration. However, for JEE Main, it's sufficient to understand that the field isn't uniform and increases as you move closer to the ends of the rod.
Perpendicular line passing through one end of the rod:
The field along this line increases initially and then reaches a constant value due to the cancellation of effects from opposite sides of the rod beyond a certain point.
The gravitational field due to a semi-infinite rod on its axis follows the same principle as a finite rod but only for the half-length portion.
The gravitational field due to an infinite rod at any point is constant along its axis and has a magnitude of:
g = 2Gλ
where λ (lambda) is the linear mass density of the rod (mass per unit length).
On the axis of a ring:
The gravitational field of a ring on its axis can be calculated using integration, but a qualitative understanding suffices for JEE Main. The field varies depending on the distance from the center of the ring.
At the center of the ring:
The gravitational field due to a ring at its center is zero due to symmetry.
The gravitational field due to a solid sphere:
The gravitational field due to a spherical shell:
Remember, for JEE Main, a qualitative understanding of the behavior of the field for these objects is sufficient. Focus on practicing problems involving calculations for simpler cases like point masses and spheres.
The acceleration due to gravity (g) experienced by an object isn't constant throughout Earth. It weakens with increasing distance from Earth's center and strengthens with decreasing distance. Here's how it varies:
As you move farther away from Earth's surface (increasing height), the acceleration due to gravity weakens. This is because the distance (r) in the formula g = G * M / r^2 increases, resulting in a smaller g value (G and M being constants for Earth).
The situation is a bit more nuanced when considering depth below the surface. Here's the breakdown:
The change in g with depth or height for practical applications near Earth's surface is minimal. For most JEE Main problems, you can assume a constant value for g (around 9.8 m/s²). However, it's crucial to understand the underlying concept for deeper understanding.
While Earth's rotation and its shape/size do affect the experienced acceleration due to gravity (g) slightly, these effects are usually negligible for most JEE Main problems. Here's a breakdown:
Earth isn't a perfect sphere; it bulges slightly at the equator due to its rotation. This creates a centrifugal force (outward force) that acts opposite to gravity.
Earth's oblate spheroid shape (bulging equator) and its overall size play a role in determining g.
Both the rotation and the shape contribute to a net decrease in g at the equator compared to the poles. However, this difference is very small. The value of g at the equator is around 9.78 m/s², while at the poles, it's about 9.83 m/s².
Gravitational potential energy (PE) is a form of energy possessed by an object due to its position in a gravitational field. It represents the potential an object has to convert that energy into kinetic energy as it moves within the field.
Imagine lifting a book off the ground. You're working against the gravitational pull of Earth. The higher you lift the book, the more work you do, and the more potential energy the book gains. If you let go, the gravitational force pulls the book down, converting its gravitational potential energy into kinetic energy of motion.
There are two main ways to categorize gravitational PE:
This refers to the gravitational potential energy an object possesses due to its interaction with another massive object, typically Earth. It's the type of PE discussed previously, where lifting an object increases its PE relative to Earth's surface.
This concept applies to objects with significant size and mass distribution, like planets or stars. Due to the gravitational attraction between different parts of the object itself, it can possess some internal gravitational PE. However, calculating self PE is complex and often not required for JEE Main.
Formula for Interaction PE:
The gravitational PE (PE) of an object due to its interaction with another massive object is given by:
PE = m g h
where:
Understanding the Formula:
Gravitational potential (V) is closely related to gravitational potential energy (PE) but offers a slightly different perspective. It describes the work done per unit mass required to move an object from a reference point (often infinity) to a specific point within a gravitational field.
Imagine a point in space within Earth's gravitational field. The gravitational potential at that point tells you how much work you'd need to do (per unit mass) to bring an object from a very far distance (considered infinity for simplicity) to that specific point, entirely against the gravitational pull.
Key Points:
The gravitational potential due to various objects can be calculated using the following general formula:
V = - G * (M / r)
where:
The negative sign indicates that work needs to be done against the attractive gravitational force.
Let's explore the potential for some specific objects:
For a point mass, the formula applies directly:
V = - G * (M / r)
The gravitational potential inside a uniform solid sphere increases linearly with distance from the center. This is because as you move inwards, the amount of mass attracting you increases, requiring more work per unit mass to overcome gravity.
The interesting aspect of a spherical shell is that the gravitational potential due to a uniform spherical shell at any point outside the shell is the same as that of a point mass with all its mass concentrated at the center. This applies due to the symmetrical cancellation of gravitational effects from opposite sides of the shell. However, the potential inside the shell is zero.
Co-Founder & Senior Physics Faculty
B.Tech. IIT-Delhi (2007)
IIT Delhi. 15+ Years of experience as Physics Faculty and Author
