Chapter 7: Gravitation Short Questions | physics Class 11 CBSE

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Chapter 7: Gravitation Short Questions | physics Class 11 CBSE | ByteNotes.in

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Question 1

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State Kepler's law of orbits and explain how it differs from the Copernican model.

Kepler's law of orbits states that all planets move in elliptical orbits with the Sun situated at one of the two foci of the ellipse.
The Copernican model assumed that planets move in perfect circular orbits around the Sun, which is a special case of an ellipse.
Kepler's law was a significant departure because it correctly described the observed non-uniform speeds of planets at different points in their orbits.

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Question 2

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Explain how Kepler's law of areas is connected to conservation of angular momentum.

Kepler's law of areas states that the line joining a planet to the Sun sweeps equal areas in equal intervals of time, expressed as ΔA/Δt = L/(2m) = constant.
Gravity is a central force directed along the line joining the Sun and the planet, so the torque on the planet about the Sun is zero, keeping angular momentum L constant.
Since L is constant and ΔA/Δt = L/(2m), the areal velocity is constant, which is precisely the law of areas.

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Question 3

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State Kepler's law of periods and verify it with the data given for Earth (a = 15.0 × 10¹⁰ m, T = 1 year).

Kepler's law of periods states that T² ∝ a³, i.e., the square of the orbital period is proportional to the cube of the semi-major axis.
For Earth: Q = T²/a³ = (1)²/(15.0)³ × 10⁻³⁰ = 2.96 × 10⁻³⁴ y² m⁻³, which matches the nearly constant value observed for all planets.
The law implies that planets farther from the Sun have larger orbital periods; for instance, Neptune with a = 450 × 10¹⁰ m has T = 165 years.

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Question 4

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Write Newton's Universal Law of Gravitation in vector form and identify each symbol.

In vector form: F = −G(m₁m₂/r²)r̂, where r̂ is the unit vector from m₁ to m₂ and the negative sign indicates the force is attractive (along −r̂).
G is the universal gravitational constant (6.67 × 10⁻¹¹ N m²/kg²), m₁ and m₂ are the masses, and r is the distance between them.
The force F₁₂ on body 1 due to body 2 equals −F₂₁ (Newton's third law), confirming that gravitational forces form an action-reaction pair.

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Question 5

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Why is it said that 'Cavendish weighed the Earth'? Explain the connection between his experiment and the mass of the Earth.

Cavendish measured the value of the universal gravitational constant G experimentally in 1798 using a torsion balance with lead spheres.
Once G is known, the mass of the Earth can be calculated from the relation g = GME/RE², since g and RE are already measurable quantities.
Thus, Cavendish's measurement of G made it possible to determine ME ≈ 5.97 × 10²⁴ kg, which is why his experiment is called 'weighing the Earth'.

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Question 6

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State the two results for the gravitational force on a point mass due to a uniform spherical shell.

If a point mass is situated outside a uniform spherical shell, the shell attracts it exactly as if its entire mass were concentrated at its centre.
If a point mass is situated inside a uniform spherical shell, the gravitational force on it due to the shell is zero, because forces from all parts of the shell cancel exactly.
These results are derived by vector summation of forces from all elements of the shell and rely on the symmetry and inverse-square nature of gravitation.

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Question 7

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Derive the expression for acceleration due to gravity g on the surface of the Earth.

For a mass m on the Earth's surface, the gravitational force is F = GMEm/RE², where ME is Earth's mass and RE its radius.
By Newton's second law, F = mg, so g = F/m = GME/RE².
This shows g depends only on the mass and radius of the Earth and not on the mass of the object; G = 6.67 × 10⁻¹¹ N m²/kg² gives g ≈ 9.8 m/s².

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Question 8

Quick Recall

Show that acceleration due to gravity decreases with height h above the Earth's surface.

At height h, the distance from the Earth's centre is (RE + h), so g(h) = GME/(RE + h)².
For h << RE, using the binomial approximation: g(h) ≈ g(1 − 2h/RE), showing g decreases linearly with small heights.
Since 2h/RE > 0, g(h) < g, confirming that gravitational acceleration is smaller at greater heights above the surface.

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Question 9

Quick Recall

Derive the expression for acceleration due to gravity at a depth d below Earth's surface.

At depth d, only the inner sphere of radius (RE − d) contributes to gravity; its mass is Ms = ME(RE − d)³/RE³.
Gravitational force: F(d) = GMsm/(RE − d)² = GMEm(RE − d)/RE³, so g(d) = GME(RE − d)/RE³ = g(1 − d/RE).
This shows g decreases linearly as depth d increases, reaching zero at the Earth's centre where d = RE.

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Question 10

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Where is the acceleration due to gravity maximum — on the surface, above, or below? Justify.

The acceleration due to gravity is maximum at the surface of the Earth.
Above the surface, g decreases as g(h) = g(1 − 2h/RE), and below the surface, g decreases as g(d) = g(1 − d/RE).
In both directions g is less than the surface value, so the surface is uniquely the point of maximum gravitational acceleration.

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Question 11

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Define gravitational potential energy and state the expression valid for arbitrary distances from the Earth.

Gravitational potential energy is the energy stored in a mass by virtue of its position in a gravitational field; it equals the work done in bringing the mass from infinity to that point.
For a mass m at distance r from the Earth's centre (r > RE), the gravitational potential energy is W(r) = −GMEm/r, with zero reference at infinity.
The negative sign indicates that energy must be supplied to move the mass to infinity; the closer the mass is to Earth, the more negative (lower) its potential energy.

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Question 12

Quick Recall

Distinguish between gravitational potential energy and gravitational potential.

Gravitational potential energy of a mass m at distance r is W(r) = −GMEm/r; it depends on both the mass m and the position.
Gravitational potential at a point is defined as the potential energy per unit mass: U(r) = −GME/r; it depends only on position and is a property of the field.
Gravitational potential is a scalar field concept useful for computing potential energy of any mass placed at that point by simply multiplying U(r) by m.

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Question 13

Quick Recall

Derive the escape speed from the surface of the Earth using energy conservation.

For an object of mass m projected with speed Vi from Earth's surface to reach infinity with zero final speed, energy conservation gives: ½mVi² − GMEm/RE = 0.
Solving: Vi = √(2GME/RE) = √(2gRE), which is the minimum (escape) speed.
Substituting g = 9.8 m/s² and RE = 6400 km, the escape speed from Earth is approximately 11.2 km/s.

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Chapter sections

Chapter 7: Gravitation | Short Questions

State Kepler's law of orbits and explain how it differs from the Copernican model.

Explain how Kepler's law of areas is connected to conservation of angular momentum.

State Kepler's law of periods and verify it with the data given for Earth (a = 15.0 × 10¹⁰ m, T = 1 year).

Write Newton's Universal Law of Gravitation in vector form and identify each symbol.

Why is it said that 'Cavendish weighed the Earth'? Explain the connection between his experiment and the mass of the Earth.

State the two results for the gravitational force on a point mass due to a uniform spherical shell.

Derive the expression for acceleration due to gravity g on the surface of the Earth.

Show that acceleration due to gravity decreases with height h above the Earth's surface.

Derive the expression for acceleration due to gravity at a depth d below Earth's surface.

Where is the acceleration due to gravity maximum — on the surface, above, or below? Justify.

Define gravitational potential energy and state the expression valid for arbitrary distances from the Earth.

Distinguish between gravitational potential energy and gravitational potential.

Derive the escape speed from the surface of the Earth using energy conservation.

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