Kepler's second law states that a line joining a planet to the Sun sweeps out equal areas in equal intervals of time, indicating variable speed. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more pleaseRead more
Kepler’s second law states that a line joining a planet to the Sun sweeps out equal areas in equal intervals of time, indicating variable speed. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
Kepler's first law states that planets move in elliptical orbits, and the Sun is at one focus of the ellipse. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more please visit here: https://www.tiwariacadeRead more
Kepler’s first law states that planets move in elliptical orbits, and the Sun is at one focus of the ellipse. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
Total energy, GmM/2r, which is half of the potential energy and negative due to gravitational binding. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/Read more
Total energy, GmM/2r, which is half of the potential energy and negative due to gravitational binding. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
The orbital velocity of a satellite near Earth's surface is the minimum velocity required for it to remain in a stable circular orbit. This velocity depends on the gravitational force acting as the centripetal force. The formula for orbital velocity (v) is: v = √(G * M / R) Where: v = orbital velociRead more
The orbital velocity of a satellite near Earth’s surface is the minimum velocity required for it to remain in a stable circular orbit. This velocity depends on the gravitational force acting as the centripetal force.
The formula for orbital velocity (v) is:
v = √(G * M / R)
Where:
v = orbital velocity
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
After calculation, the orbital velocity is approximately:
v ≈ 7.9 km/s (or 7,900 m/s)
This velocity is valid for satellites close to the Earth’s surface and ensures that the satellite remains in orbit without falling back to Earth or escaping into space.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
The escape velocity is the minimum velocity an object must have to break free from the gravitational pull of a planet, such as Earth, without further propulsion. The formula for escape velocity (vₑ) is: vₑ = √(2 * G * M / R) Where: vₑ = escape velocity G = gravitational constant (approximately 6.674Read more
The escape velocity is the minimum velocity an object must have to break free from the gravitational pull of a planet, such as Earth, without further propulsion.
The formula for escape velocity (vₑ) is:
vₑ = √(2 * G * M / R)
Where:
vₑ = escape velocity
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
After calculation, the escape velocity is approximately:
vₑ ≈ 11.2 km/s (or 11,200 m/s)
This means an object must travel at least 11.2 km/s to escape Earth’s gravitational influence completely.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
What is Kepler’s second law?
Kepler's second law states that a line joining a planet to the Sun sweeps out equal areas in equal intervals of time, indicating variable speed. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more pleaseRead more
Kepler’s second law states that a line joining a planet to the Sun sweeps out equal areas in equal intervals of time, indicating variable speed. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/
This law describes planetary orbits.
Kepler's first law states that planets move in elliptical orbits, and the Sun is at one focus of the ellipse. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more please visit here: https://www.tiwariacadeRead more
Kepler’s first law states that planets move in elliptical orbits, and the Sun is at one focus of the ellipse. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/
What is the total energy of a satellite in a circular orbit?
Total energy, GmM/2r, which is half of the potential energy and negative due to gravitational binding. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/Read more
Total energy, GmM/2r, which is half of the potential energy and negative due to gravitational binding. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/
What is the orbital velocity for a satellite near Earths surface?
The orbital velocity of a satellite near Earth's surface is the minimum velocity required for it to remain in a stable circular orbit. This velocity depends on the gravitational force acting as the centripetal force. The formula for orbital velocity (v) is: v = √(G * M / R) Where: v = orbital velociRead more
The orbital velocity of a satellite near Earth’s surface is the minimum velocity required for it to remain in a stable circular orbit. This velocity depends on the gravitational force acting as the centripetal force.
The formula for orbital velocity (v) is:
v = √(G * M / R)
Where:
v = orbital velocity
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
Substituting the values of G, M, and R for Earth:
v = √((6.674 × 10^-11) * (5.972 × 10^24) / 6.371 × 10^6)
After calculation, the orbital velocity is approximately:
v ≈ 7.9 km/s (or 7,900 m/s)
This velocity is valid for satellites close to the Earth’s surface and ensures that the satellite remains in orbit without falling back to Earth or escaping into space.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/
What is the escape velocity for Earth?
The escape velocity is the minimum velocity an object must have to break free from the gravitational pull of a planet, such as Earth, without further propulsion. The formula for escape velocity (vₑ) is: vₑ = √(2 * G * M / R) Where: vₑ = escape velocity G = gravitational constant (approximately 6.674Read more
The escape velocity is the minimum velocity an object must have to break free from the gravitational pull of a planet, such as Earth, without further propulsion.
The formula for escape velocity (vₑ) is:
vₑ = √(2 * G * M / R)
Where:
vₑ = escape velocity
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
Substituting the values of G, M, and R for Earth:
vₑ = √(2 * (6.674 × 10^-11) * (5.972 × 10^24) / 6.371 × 10^6)
After calculation, the escape velocity is approximately:
vₑ ≈ 11.2 km/s (or 11,200 m/s)
This means an object must travel at least 11.2 km/s to escape Earth’s gravitational influence completely.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/