The satellite and everything inside it are in free fall towards Earth, causing apparent weightlessness. 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.comRead more
The satellite and everything inside it are in free fall towards Earth, causing apparent weightlessness. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
The acceleration due to gravity (g) decreases with an increase in altitude because the distance from the center of the Earth increases. The gravitational force follows the inverse square law. The formula to calculate the value of g at a height (h) above the Earth's surface is: gₕ = g₀ * (R / (R + h)Read more
The acceleration due to gravity (g) decreases with an increase in altitude because the distance from the center of the Earth increases. The gravitational force follows the inverse square law.
The formula to calculate the value of g at a height (h) above the Earth’s surface is:
gₕ = g₀ * (R / (R + h))²
Where:
gₕ = acceleration due to gravity at height h
g₀ = acceleration due to gravity on Earth’s surface (approximately 9.8 m/s²)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
h = height above Earth’s surface
### Key Points:
1. At higher altitudes, the distance (R + h) from the Earth’s center increases, resulting in a smaller value of gₕ.
2. The decrease in g is noticeable at significant altitudes, such as those of satellites or mountain peaks, but negligible for everyday heights.
3. At very high altitudes, where h is comparable to or larger than R, the effect becomes more pronounced.
### Example:
For an altitude of 100 km (h = 100,000 m):
After calculation, gₕ is slightly less than 9.8 m/s², illustrating the decrease in g with altitude.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
The acceleration due to gravity on Earth's surface is approximately: g₀ = 9.8 m/s² This value represents the gravitational force experienced by an object per unit mass at the Earth's surface. It is derived from the formula: g₀ = G * M / R² Where: g₀ = acceleration due to gravity on Earth's surface GRead more
The acceleration due to gravity on Earth’s surface is approximately:
g₀ = 9.8 m/s²
This value represents the gravitational force experienced by an object per unit mass at the Earth’s surface. It is derived from the formula:
g₀ = G * M / R²
Where:
g₀ = acceleration due to gravity on Earth’s surface
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 these values results in the standard value of g₀ ≈ 9.8 m/s².
This value can vary slightly depending on altitude, latitude, and local geological structures.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
The gravitational force between the Earth and the Sun is primarily determined by: 1. **Mass of the Sun (Mₛ):** The Sun's mass is a key factor in the gravitational force. The larger the mass of the Sun, the stronger the gravitational pull it exerts on the Earth. 2. **Mass of the Earth (Mₑ):** The EarRead more
The gravitational force between the Earth and the Sun is primarily determined by:
1. **Mass of the Sun (Mₛ):**
The Sun’s mass is a key factor in the gravitational force. The larger the mass of the Sun, the stronger the gravitational pull it exerts on the Earth.
2. **Mass of the Earth (Mₑ):**
The Earth’s mass also contributes to the gravitational force. The force is proportional to the product of the two masses (Mₛ * Mₑ).
3. **Distance between Earth and Sun (r):**
The distance between the Earth and the Sun plays a critical role in determining the gravitational force. According to the inverse square law, the gravitational force decreases with the square of the distance (1/r²).
The gravitational force is mathematically given by:
F = G * (Mₛ * Mₑ) / r²
Where:
F = gravitational force
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
Mₛ = mass of the Sun (approximately 1.989 × 10^30 kg)
Mₑ = mass of the Earth (approximately 5.972 × 10^24 kg)
r = average distance between Earth and Sun (approximately 1.496 × 10^11 m)
Key Point:
The two most significant factors in determining the gravitational force are the masses of the Earth and the Sun, and the distance between them.
This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
Inside a uniform spherical shell, gravitational field is zero due to symmetry and the shell theorem. 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/ncRead more
Inside a uniform spherical shell, gravitational field is zero due to symmetry and the shell theorem. This question related to Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Give answer according to your understanding.
Why do astronauts experience weightlessness in a satellite?
The satellite and everything inside it are in free fall towards Earth, causing apparent weightlessness. 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.comRead more
The satellite and everything inside it are in free fall towards Earth, causing apparent weightlessness. 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/
How does 𝑔 vary with altitude?
The acceleration due to gravity (g) decreases with an increase in altitude because the distance from the center of the Earth increases. The gravitational force follows the inverse square law. The formula to calculate the value of g at a height (h) above the Earth's surface is: gₕ = g₀ * (R / (R + h)Read more
The acceleration due to gravity (g) decreases with an increase in altitude because the distance from the center of the Earth increases. The gravitational force follows the inverse square law.
The formula to calculate the value of g at a height (h) above the Earth’s surface is:
gₕ = g₀ * (R / (R + h))²
Where:
gₕ = acceleration due to gravity at height h
g₀ = acceleration due to gravity on Earth’s surface (approximately 9.8 m/s²)
R = radius of the Earth (approximately 6,371 km or 6.371 × 10^6 m)
h = height above Earth’s surface
### Key Points:
1. At higher altitudes, the distance (R + h) from the Earth’s center increases, resulting in a smaller value of gₕ.
2. The decrease in g is noticeable at significant altitudes, such as those of satellites or mountain peaks, but negligible for everyday heights.
3. At very high altitudes, where h is comparable to or larger than R, the effect becomes more pronounced.
### Example:
For an altitude of 100 km (h = 100,000 m):
gₕ = 9.8 * (6.371 × 10^6 / (6.371 × 10^6 + 100,000))²
After calculation, gₕ is slightly less than 9.8 m/s², illustrating the decrease in g with altitude.
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 value of acceleration due to gravity on Earths surface?
The acceleration due to gravity on Earth's surface is approximately: g₀ = 9.8 m/s² This value represents the gravitational force experienced by an object per unit mass at the Earth's surface. It is derived from the formula: g₀ = G * M / R² Where: g₀ = acceleration due to gravity on Earth's surface GRead more
The acceleration due to gravity on Earth’s surface is approximately:
g₀ = 9.8 m/s²
This value represents the gravitational force experienced by an object per unit mass at the Earth’s surface. It is derived from the formula:
g₀ = G * M / R²
Where:
g₀ = acceleration due to gravity on Earth’s surface
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 these values results in the standard value of g₀ ≈ 9.8 m/s².
This value can vary slightly depending on altitude, latitude, and local geological structures.
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/
The gravitational force between Earth and Sun keeps Earth in its orbit. What primarily determines this force?
The gravitational force between the Earth and the Sun is primarily determined by: 1. **Mass of the Sun (Mₛ):** The Sun's mass is a key factor in the gravitational force. The larger the mass of the Sun, the stronger the gravitational pull it exerts on the Earth. 2. **Mass of the Earth (Mₑ):** The EarRead more
The gravitational force between the Earth and the Sun is primarily determined by:
1. **Mass of the Sun (Mₛ):**
The Sun’s mass is a key factor in the gravitational force. The larger the mass of the Sun, the stronger the gravitational pull it exerts on the Earth.
2. **Mass of the Earth (Mₑ):**
The Earth’s mass also contributes to the gravitational force. The force is proportional to the product of the two masses (Mₛ * Mₑ).
3. **Distance between Earth and Sun (r):**
The distance between the Earth and the Sun plays a critical role in determining the gravitational force. According to the inverse square law, the gravitational force decreases with the square of the distance (1/r²).
The gravitational force is mathematically given by:
F = G * (Mₛ * Mₑ) / r²
Where:
F = gravitational force
G = gravitational constant (approximately 6.674 × 10^-11 N m²/kg²)
Mₛ = mass of the Sun (approximately 1.989 × 10^30 kg)
Mₑ = mass of the Earth (approximately 5.972 × 10^24 kg)
r = average distance between Earth and Sun (approximately 1.496 × 10^11 m)
Key Point:
The two most significant factors in determining the gravitational force are the masses of the Earth and the Sun, and the distance between them.
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 gravitational field inside a uniform spherical shell?
Inside a uniform spherical shell, gravitational field is zero due to symmetry and the shell theorem. 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/ncRead more
Inside a uniform spherical shell, gravitational field is zero due to symmetry and the shell theorem. 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/