4 Most SAQ’s of Gravitation Chapter in Inter 1st Year Physics (TS/AP)

4 Marks

SAQ-1 : What is geostationary satellite ? State its uses.

Introduction

A geostationary satellite is a satellite that orbits the Earth at the same rotational rate as the Earth on its axis. This unique orbit results in the satellite appearing stationary relative to a fixed point on the Earth’s surface. Positioned approximately 35,786 kilometers above the equator, these satellites provide continuous and stable coverage of a specific region on Earth.

Uses of Geostationary Satellite

1. Telecommunications and Broadcasting: Geostationary satellites are crucial for telecommunications and broadcasting. They enable long-distance communication, including television broadcasting, internet services, phone calls, and data transmission. Their fixed position relative to the Earth allows for stable connections without needing to reorient antennas.
2. Weather Monitoring: These satellites play a vital role in weather monitoring and forecasting. They provide real-time images and data on weather patterns, cloud movements, and atmospheric conditions, aiding meteorologists in tracking storms and predicting weather changes.
3. Navigation and Global Positioning System (GPS): While geostationary satellites are not typically used for navigation, they complement a network of low Earth orbit (LEO) satellites for systems like GPS, enabling accurate positioning, navigation, and timing services.
4. Earth Observation: Geostationary satellites are powerful tools for Earth observation, capturing high-resolution images of the Earth’s surface. They monitor environmental changes and provide data for applications in agriculture, forestry, urban planning, disaster management, and environmental studies.
5. Satellite TV and Radio Broadcasting: A common use of geostationary satellites is for direct-to-home (DTH) satellite television and radio broadcasting, broadcasting TV channels and radio signals directly to homes and vehicles.
6. Internet Connectivity in Remote Areas: These satellites provide internet connectivity in remote and underserved areas. They facilitate broadband internet access and help bridge the digital divide in remote regions.

Summary

Geostationary Satellites orbit the Earth in sync with the Earth’s rotation, appearing stationary relative to the Earth’s surface. Their applications are diverse, encompassing telecommunications, broadcasting, weather monitoring, Earth observation, satellite TV, radio broadcasting, and internet connectivity in remote areas. Their unique capability to remain fixed above a specific location on Earth makes them invaluable for a wide range of essential services and technologies.

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SAQ-2 : What is Orbital velocity? Obtain an expression for it.

Introduction

Orbital Velocity is the velocity required for an object to enter and maintain a stable orbit around a celestial body, such as a planet or a star. It is the speed at which the gravitational pull of the body and the inertia of the object are balanced, allowing the object to follow a circular path without falling into the body or escaping into space.

Derivation of the Expression for Orbital Velocity

1. Gravitational Force and Centripetal Force:
   • In a stable orbit, the gravitational force acting on the orbiting object provides the necessary centripetal force to keep it in orbit.
   • The gravitational force is given by $$F_{\text{gravity}} = \frac{GmM}{r^2}$$ where G is the gravitational constant, m is the mass of the orbiting object, M is the mass of the central body, and r is the radius of the orbit.
2. Centripetal Force for Circular Motion: The centripetal force required for circular motion is given by $$F_{\text{centripetal}} = \frac{mv^2}{r}$$ where v is the orbital velocity.
3. Equating Gravitational and Centripetal Forces: Setting F_gravity​ = F_{centripetal​} leads to the equation $$\frac{GmM}{r^2} = \frac{mv^2}{r}$$
4. Solving for Orbital Velocity:
   • Simplifying the equation, we get $$G\frac{M}{r} = v^2$$
   • Therefore, the orbital velocity v is obtained as $$v = \sqrt{\frac{GM}{r}}$$

Summary

Orbital Velocity is crucial for an object to maintain a stable orbit around a celestial body. The expression for orbital velocity is derived by equating the gravitational force to the centripetal force required for circular motion, resulting in the formula $$v = \sqrt{\frac{GM}{r}}$$ This formula highlights the relationship between the mass of the central body, the radius of the orbit, and the velocity required to maintain the orbit.

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SAQ-3 : What is escape velocity? Obtain an expression for it.

Introduction

Escape Velocity is the minimum velocity required for an object to break free from the gravitational pull of a celestial body, such as a planet or a moon, without further propulsion. It is the speed needed to overcome the gravitational potential energy and escape into space.

Derivation of the Expression for Escape Velocity

1. Gravitational Potential Energy: The gravitational potential energy of an object at a distance r from the center of a celestial body (of mass M) is given by $$U = -\frac{GmM}{r}$$ where G is the gravitational constant, and m is the mass of the object.
2. Kinetic Energy for Escape: To escape the gravitational pull, an object needs sufficient kinetic energy equal to its gravitational potential energy at that point. The kinetic energy is given by $$K = \frac{1}{2}mv^2$$ where v is the escape velocity.
3. Equating Kinetic and Potential Energies: Setting the kinetic energy equal to the magnitude of the gravitational potential energy, we get $$\frac{1}{2}mv^2 = \frac{GmM}{r}$$
4. Solving for Escape Velocity: Solving for v, the escape velocity is obtained as $$v = \sqrt{\frac{2GM}{r}}$$

Summary

Escape Velocity is the critical velocity required for an object to escape a celestial body’s gravitational field without additional propulsion. The formula for escape velocity is derived by equating the kinetic energy needed to escape with the gravitational potential energy, resulting in the expression $$v = \sqrt{\frac{2GM}{r}}$$ This formula is pivotal in understanding the energy requirements for space missions and celestial mechanics.

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SAQ-4 : State Kepler’s Laws of Planetary motion.

Introduction

Johannes Kepler, a notable German mathematician and astronomer, formulated three fundamental laws that describe the motion of planets around the Sun. These laws were pivotal in revolutionizing our understanding of the solar system and significantly contributed to the development of Isaac Newton’s law of universal gravitation.

1. Kepler’s First Law (Law of Ellipses): According to Kepler’s First Law, each planet orbits the Sun in an elliptical path, with the Sun located at one of the two foci of the ellipse. This law highlights that the orbits of planets are ellipses rather than perfect circles, with the Sun occupying one of the focal points.
2. Kepler’s Second Law (Law of Equal Areas): Kepler’s Second Law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time intervals. This implies that a planet’s orbital speed varies; it moves faster when closer to the Sun (perihelion) and slower when farther away (aphelion).
3. Kepler’s Third Law (Law of Harmonies): The Third Law, also known as the Law of Harmonies, establishes that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit. Mathematically, this is represented as T² ∝ a³. It indicates that the farther a planet is from the Sun, the longer its orbital period.

Summary

Kepler’s laws significantly advanced the heliocentric model of the solar system proposed by Nicolaus Copernicus. They provided empirical evidence supporting the elliptical motion of planets around the Sun, challenging the earlier belief in perfect circular orbits. These laws laid the foundation for future discoveries in celestial mechanics and were instrumental in formulating Newton’s law of universal gravitation.