Physics · Class 11 Science
Ch 7Gravitation — Class 11 Physics, concept-first.
For most of human history, the motion of the planets, the Moon and the Sun across the sky was described only in terms of how they appeared to move, without any single physical principle explaining why they moved that way.
Key concepts
Hover a concept to preview it and jump to its most relevant Q&A.
Newton's Law of Universal Gravitation
Every particle of matter attracts every other particle with a force directed along the line joining them. For two point masses and separated by a distance , the magnitude of the attraction is
Most relevant Q&A
- State Newton's universal law of gravitation in words and as an equation. Identify the SI unit and the accepted value of the universal gravit…Free
- Two point masses of $50\ \text{kg}$ and $80\ \text{kg}$ are placed with their centres $2\ \text{m}$ apart. Find the gravitational force of a…Free
- Starting from Newton's universal law of gravitation, derive the expression $g = GM/R^2$ for the acceleration due to gravity at the Earth's s…Free
- Calculate the gravitational force of attraction between the Earth ($M = 6 \times 10^{24}\ \text{kg}$) and the Moon ($m = 7.4 \times 10^{22}\…Free
In previous exams
How often this chapter’s concepts have been examined — real appearance data, never estimated.
Chapter contents
The NCERT structure, section by section. Open a section to see its questions, then read the concept-first solution.
Introduction
For most of human history, the motion of the planets, the Moon and the Sun across the sky was described only in terms of how they appeared to move, without any single physical principle explaining why…
Kepler's Laws of Planetary Motion
Johannes Kepler analysed decades of the astronomer Tycho Brahe's painstakingly accurate naked-eye observations of the planet Mars and, from them, worked out three precise geometrical laws that every p…
Newton's Universal Law of Gravitation
Newton's universal law of gravitation states: every particle of matter in the universe attracts every other particle with a force whose magnitude is directly proportional to the product of their two m…
The Universal Gravitational Constant G
The constant appearing in Newton's law of gravitation, , is called the universal gravitational constant. Its currently accepted value is
Acceleration Due to Gravity of the Earth
Consider an object of mass resting at the Earth's surface, a distance (the Earth's own radius) from the Earth's centre, and treat the Earth, for this purpose, as though its entire mass were concentrat…
Variation of g with Altitude and Depth
Variation with altitude. Consider a point at height above the Earth's surface, so that its distance from the Earth's centre is rather than . Applying at this new distance,
Variation of g Due to Rotation of the Earth
The Earth spins on its own axis once approximately every hours, and this rotation provides a third reason -- entirely separate from altitude or depth -- why the acceleration due to gravity actually me…
Gravitational Potential Energy
Near the Earth's surface, over small height changes , the gravitational potential energy gained by a mass is given by the familiar formula (measured relative to the ground, at ), because itself can sa…
Gravitational Potential
Gravitational potential at a point in a gravitational field is defined as the gravitational potential energy PER UNIT MASS that a small test mass would have if placed at that point:
Escape Velocity
Escape velocity is defined as the minimum speed with which an object must be launched from a planet's surface, straight upward (with no further propulsion after launch, and ignoring air resistance), s…
Orbital Velocity of a Satellite
Consider a satellite of mass moving in a circular orbit of radius (measured from the Earth's centre) around the Earth.
Total Energy of an Orbiting Satellite
A satellite in a circular orbit of radius possesses both kinetic energy (from its orbital motion) and gravitational potential energy (from its position in the Earth's gravitational field); adding the…
Geostationary Satellite
A geostationary satellite is an Earth satellite that appears completely stationary in the sky when viewed from any fixed point on the ground -- an enormously useful property for communication and weat…
Summary
This chapter built up WBCHSE Unit 6's account of gravitation in the order the syllabus lists it. Kepler's three laws (Section 7.2) described planetary motion purely geometrically -- elliptical orbits…
Sample & Board Papers
Sample papers and previous-year board questions for this subject.
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- Q1Keeping the mass fixed, if the radius of the earth is halved, the acceleration due to gravity at any place will be (a) half of the original…Preview
- Q2Define velocity of escape. Find an expression of the velocity of escape of a body projected from the surface of the earth. **OR** Show that…Preview
- Q3Define acceleration due to gravity. Deduce an expression of the acceleration due to gravity at a place on the surface of the earth due to it…Preview
More questions
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- Example 1State Newton's universal law of gravitation in words and as an equation. Identify the SI unit and the accepted value of the universal gravit…Free
- Example 2Two point masses of $50\ \text{kg}$ and $80\ \text{kg}$ are placed with their centres $2\ \text{m}$ apart. Find the gravitational force of a…Free
- Example 3Find the acceleration due to gravity at a height above the Earth's surface equal to the Earth's own radius $R$. Take $g = 9.8\ \text{m/s}^2$…Free
- Example 4Find the acceleration due to gravity at a depth equal to half the Earth's radius, i.e. $d = R/2$, below the Earth's surface. Take $g = 9.8\…Preview
- Example 5Calculate the escape velocity from the Earth's surface, given $g = 9.8\ \text{m/s}^2$ and $R = 6.4 \times 10^6\ \text{m}$.Preview
- Example 6A satellite revolves around the Earth in a circular orbit at a height of $300\ \text{km}$ above the surface. Taking $GM = 4.00 \times 10^{14…Preview
- Example 7A geostationary satellite is to be placed in an equatorial orbit with a time period of exactly $24\ \text{h}$. Using $GM = 4.00 \times 10^{1…Preview
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- Q8State Kepler's three laws of planetary motion.Free
- Q9Using Kepler's second law (the law of areas), explain why a planet moves fastest when it is closest to the Sun (at perihelion) and slowest w…Free
- Q10Starting from Newton's universal law of gravitation, derive the expression $g = GM/R^2$ for the acceleration due to gravity at the Earth's s…Free
- Q11Derive an expression for the acceleration due to gravity at a depth $d$ below the Earth's surface, assuming the Earth to be a uniform sphere…Preview
- Q12Derive an expression for the effective acceleration due to gravity $g'$ at a place of latitude $\lambda$, taking into account the Earth's ro…Preview
- Q13Starting from the work done in bringing a point mass $m$ from infinity to a distance $r$ from a point mass $M$, derive the expression for gr…Preview
- Q14Define gravitational potential at a point in a gravitational field. Derive the expression $V(r) = -GM/r$ for the potential at a distance $r$…Preview
- Q15Define escape velocity and orbital velocity. For a satellite in a circular orbit very close to a planet's surface, derive the relation $v_e…Preview
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- Q16Calculate the gravitational force of attraction between the Earth ($M = 6 \times 10^{24}\ \text{kg}$) and the Moon ($m = 7.4 \times 10^{22}\…Free
- Q17Find the value of $g$ at a height of $1600\ \text{km}$ above the Earth's surface. Take $g = 9.8\ \text{m/s}^2$ and $R = 6400\ \text{km}$.Free
- Q18At what depth below the Earth's surface does the acceleration due to gravity reduce to exactly $50\%$ of its value at the surface? Take $R =…Free
- Q19Estimate the reduction in the apparent value of $g$ at the equator, caused purely by the Earth's rotation about its own axis. Take $R = 6.4…Preview
- Q20A body weighs $63\ \text{N}$ on the Earth's surface. What would it weigh at a height equal to half the Earth's radius, i.e. $h = R/2$, above…Preview
- Q21Calculate the escape velocity from the surface of the Moon, given the Moon's radius as $1.74 \times 10^{6}\ \text{m}$ and its mass as $7.4 \…Preview
- Q22A geostationary satellite has a time period of exactly $24\ \text{h}$. Taking $M = 6 \times 10^{24}\ \text{kg}$, $R = 6400\ \text{km}$ and $…Preview
- Q23A satellite of mass $1000\ \text{kg}$ orbits the Earth in a circular path of radius $6700\ \text{km}$ (i.e. at a height of $300\ \text{km}$)…Preview