Chapter 7: Gravitation Application Questions | physics Class 11 CBSE

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

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 1

Applied Concept

A spacecraft is launched from Earth to Mars. With reference to Kepler's laws, explain why the spacecraft must travel in an elliptical transfer orbit and how mission planners choose the launch window.

By Kepler's first law, any orbit around the Sun (including a transfer trajectory) must be elliptical with the Sun at one focus; the spacecraft's transfer orbit has perihelion at Earth's orbit and aphelion at Mars's orbit.
The spacecraft must be launched when Mars is positioned such that it arrives at the aphelion point just as the spacecraft completes half of its elliptical transfer orbit — this alignment defines the launch window.
By Kepler's third law, the transfer orbit period can be calculated, and launch windows repeat whenever Earth and Mars are correctly aligned (approximately every 26 months).
Kepler's second law implies the spacecraft moves slowest at aphelion (near Mars), so the required velocity change at arrival is minimised when approaching Mars at aphelion.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A mining company proposes to drill a tunnel to the Earth's centre. With reference to the variation of g with depth, analyse how the gravitational acceleration and weight of a drill would change during descent.

As the drill descends to depth d, only the inner sphere of radius (RE − d) contributes to gravity; g(d) = g(1 − d/RE), decreasing linearly with depth.
At half the Earth's radius (d = RE/2), g(d) = g/2 ≈ 4.9 m/s², so the drill's apparent weight halves; at the Earth's centre, g = 0 and the drill becomes weightless.
The drill would experience maximum weight at the surface and continuously lighter conditions as it descends, making deep drilling progressively less energy-intensive in terms of gravitational load.
This analysis assumes uniform Earth density; in reality Earth is denser at the core, so the actual variation of g is different from the simple linear model, but the chapter's uniform-density formula provides the baseline.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A satellite communication company wants to place satellites at geostationary orbit so they appear stationary over a location. Using the period formula for satellites, explain the conditions and calculate the required orbital height (use T = 24 h, ME = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m²/kg², RE = 6400 km).

For a geostationary satellite, the orbital period must equal Earth's rotation period: T = 24 hours = 86400 s, and the orbit must be equatorial to remain stationary over the same point.
From T² = (4π²/GME)(RE + h)³, solving for (RE + h): (RE + h)³ = GME T²/(4π²) = (6.67×10⁻¹¹ × 6×10²⁴ × (86400)²)/(4π²).
Computing: (RE + h)³ ≈ 7.57 × 10²² m³, giving RE + h ≈ 4.23 × 10⁷ m, so h ≈ 35,900 km above Earth's surface.
At this altitude, the satellite's orbital period exactly matches Earth's rotation, making it appear stationary — this is the basis for all geostationary communication and weather satellites.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to the concept of escape speed, explain why a rocket bound for interplanetary space requires multiple stages, and why the escape speed from Earth is approximately 11.2 km/s.

Escape speed from Earth's surface is ve = √(2gRE) = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s, derived by setting total energy to zero at infinity using energy conservation.
A single-stage rocket would need to carry enormous fuel to reach this speed; by jettisoning empty fuel tanks in stages, a multi-stage rocket reduces the mass being accelerated, greatly improving efficiency.
Since escape speed is independent of the rocket's mass but depends on Earth's mass and radius, the 11.2 km/s threshold applies regardless of rocket size.
After escaping Earth's gravity, the spacecraft must also overcome the Sun's gravity for interplanetary missions, requiring additional speed beyond Earth's escape speed.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

An astronaut in the International Space Station (ISS, orbiting at ~400 km) reports feeling weightless. Explain this using Newton's law of gravitation and the concept of free fall.

At 400 km altitude, g ≈ GME/(RE + 400 km)² ≈ 8.7 m/s² — gravity is still significant and not negligible, so weightlessness cannot be due to absence of gravity.
Both the ISS and the astronaut are in free fall toward Earth with the same acceleration 8.7 m/s²; the ISS's horizontal velocity is just enough to keep missing the Earth as it falls.
Since the astronaut and the station share the same acceleration, there is no contact force (normal reaction) between them — this absence of contact force is experienced as weightlessness.
Weightlessness in orbit is identical in character to weightlessness during free fall in a lift whose cable has snapped; it is not due to gravity being zero but to being in free fall.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

A rescue team uses a gravitational survey instrument (gravimeter) to detect underground cavities. With reference to the shell theorem, explain the physical principle behind such measurements.

In uniform rock, each point at the surface has the same g value determined by the total mass below (shell theorem: only mass below contributes, shells above cancel).
An underground cavity is a region of missing mass; at the surface directly above the cavity, the gravitational pull is slightly less than normal because rock mass is replaced by air or void.
A gravimeter detects this reduction in g — the cavity removes mass that would otherwise have contributed to the downward gravitational pull at the surface.
The shell theorem ensures that only the missing mass directly below (in the relevant sphere of influence) affects the surface gravity reading, allowing cavity depth and size to be estimated from the magnitude and spatial distribution of g anomalies.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

India's ISRO launches a remote sensing satellite into a polar orbit at 600 km altitude. Using satellite dynamics, calculate its approximate orbital speed and time period, and explain why it covers the entire Earth.

Orbital speed: V = √[GME/(RE + h)] = √[9.8 × (6.4 × 10⁶)²/(7.0 × 10⁶)] ≈ √(5.74 × 10⁷) ≈ 7.58 km/s.
Time period: T = 2π(RE + h)/V = 2π × 7.0 × 10⁶ / 7580 ≈ 5800 s ≈ 97 minutes.
A polar orbit passes over both poles; as the satellite completes each ~97-minute orbit, the Earth rotates about 24° beneath it, so successive passes cover adjacent strips of the surface.
Within approximately 14 orbits per day, the satellite scans the entire Earth, making polar orbits ideal for remote sensing, weather observation, and reconnaissance.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

Using the concept of gravitational potential energy, explain why it is more economical for a spacecraft to perform a gravitational slingshot manoeuvre around a planet rather than using rocket thrust alone to gain speed.

In a gravitational slingshot, the spacecraft approaches a massive planet and is deflected; from the planet's reference frame, the spacecraft enters and leaves at the same speed (elastic encounter), but from the Sun's frame, it can gain or lose speed.
The spacecraft gains kinetic energy by converting some of the planet's orbital kinetic energy through tidal gravitational interaction — gravitational potential energy is temporarily borrowed and returned, but the spacecraft ends up in a higher-energy trajectory.
Rocket thrust requires carrying propellant whose mass reduces the rocket's efficiency; every kilogram of fuel burned must also carry earlier fuel to orbit, making deep-space travel fuel-expensive.
A slingshot uses the planet's existing gravitational field and orbital energy at no propellant cost, effectively giving 'free' delta-v, which can reduce mission cost and travel time significantly.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A geophysics student notes that a pendulum clock runs slower at higher altitudes and faster in deep mines compared to at the Earth's surface. Explain this observation using the variation of g.

The time period of a simple pendulum is T = 2π√(L/g); since T ∝ 1/√g, a smaller g means a longer period and a slower-running clock.
At higher altitudes, g decreases as g(h) ≈ g(1 − 2h/RE), so the pendulum's period increases and the clock runs slow.
In a deep mine at depth d, g(d) = g(1 − d/RE) also decreases (though more slowly than above), so the clock also runs slower in mines compared to at the surface.
The surface is the unique point of maximum g, where the clock runs fastest; both going up and going down make the pendulum slower, consistent with the chapter's result that g is maximum at Earth's surface.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to the law of areas, explain why a comet moves very fast near the Sun and very slowly far away, and whether its angular momentum and total energy are conserved.

By Kepler's second law, the line joining the comet to the Sun sweeps equal areas in equal times. Near the Sun (small r), to sweep the same area as far away (large r), the comet must move much faster.
Angular momentum L = mr × v is conserved because gravity is a central force with zero torque; so as r decreases near perihelion, v must increase to keep L = mrv constant.
Total mechanical energy (KE + PE) is also conserved throughout the orbit since gravity is a conservative force; as the comet falls toward the Sun, it loses PE and gains KE.
Neither linear momentum nor individual KE or PE values are constant — only angular momentum and total mechanical energy are conserved for an isolated gravitational system.

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Chapter sections

Chapter 7: Gravitation | Application Questions

A spacecraft is launched from Earth to Mars. With reference to Kepler's laws, explain why the spacecraft must travel in an elliptical transfer orbit and how mission planners choose the launch window.

A mining company proposes to drill a tunnel to the Earth's centre. With reference to the variation of g with depth, analyse how the gravitational acceleration and weight of a drill would change during descent.

With reference to the concept of escape speed, explain why a rocket bound for interplanetary space requires multiple stages, and why the escape speed from Earth is approximately 11.2 km/s.

An astronaut in the International Space Station (ISS, orbiting at ~400 km) reports feeling weightless. Explain this using Newton's law of gravitation and the concept of free fall.

A rescue team uses a gravitational survey instrument (gravimeter) to detect underground cavities. With reference to the shell theorem, explain the physical principle behind such measurements.

India's ISRO launches a remote sensing satellite into a polar orbit at 600 km altitude. Using satellite dynamics, calculate its approximate orbital speed and time period, and explain why it covers the entire Earth.

Using the concept of gravitational potential energy, explain why it is more economical for a spacecraft to perform a gravitational slingshot manoeuvre around a planet rather than using rocket thrust alone to gain speed.

A geophysics student notes that a pendulum clock runs slower at higher altitudes and faster in deep mines compared to at the Earth's surface. Explain this observation using the variation of g.

With reference to the law of areas, explain why a comet moves very fast near the Sun and very slowly far away, and whether its angular momentum and total energy are conserved.

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