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.

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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:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/

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:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-7/