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What happens to the orbital period of a satellite if its orbital radius increases? Description: Consider Kepler’s third law.
By Kepler’s third law, T² ∝ r³. As the orbital radius increases, the period also increases. 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-solutRead more
By Kepler’s third law, T² ∝ r³. As the orbital radius increases, the period also increases. 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:
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A bullet of mass 10 g leaves a rifle at an initial velocity of 100 m/s and strikes the earth at the same level with a velocity of 500 m/s. The work done in joule overcoming the resistance of air will be
To calculate the work done overcoming the resistance of air, we will need to determine the change in kinetic energy. Step 1: Write down the formula for work done The work done by air resistance is equal to the loss in the kinetic energy of the bullet. The formula for kinetic energy is: K.E. = (1/2)Read more
To calculate the work done overcoming the resistance of air, we will need to determine the change in kinetic energy.
Step 1: Write down the formula for work done
The work done by air resistance is equal to the loss in the kinetic energy of the bullet. The formula for kinetic energy is:
K.E. = (1/2) m v²
Where:
– m = mass of the bullet = 10 g = 0.01 kg
– v = velocity of the bullet (initial and final)
Step 2: Calculate the initial and final kinetic energies
Initial velocity, u = 100 m/s
Final velocity, v = 500 m/s
Initial kinetic energy:
K.E.₁ = (1/2) m u²
K.E.₁ = (1/2) × 0.01 × (100)²
K.E.₁ = 0.005 × 10000
K.E.₁ = 50 J
Final kinetic energy:
K.E.₂ = (1/2) m v²
K.E.₂ = (1/2) × 0.01 × (500)²
K.E.₂ = 0.005 × 250000
K.E.₂ = 1250 J
Step 3: Find the work done
The work done in overcoming air resistance is the difference in kinetic energy:
Work done = K.E.₂ − K.E.₁
Work done = 1250 − 50
Work done = 1200 J
The work done in overcoming the resistance of air is 1200 J.
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How does escape velocity depend on the mass of the planet?
The escape velocity (v_e) is the minimum velocity an object must have in order to break free from a planet's gravitational influence without further propulsion. It is derived from the balance between the object's kinetic energy and the planet's gravitational potential energy. The formula for escapeRead more
The escape velocity (v_e) is the minimum velocity an object must have in order to break free from a planet’s gravitational influence without further propulsion. It is derived from the balance between the object’s kinetic energy and the planet’s gravitational potential energy.
The formula for escape velocity is:
v_e = √(2GM / R)
Where:
– v_e is the escape velocity,
– G is the gravitational constant (6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²),
– M is the mass of the planet,
– R is the radius of the planet.
From this formula, we can see that escape velocity depends on the mass of the planet (M) in a square root relationship. Specifically, the escape velocity increases with the square root of the mass. This means that for a planet with greater mass, the escape velocity will be higher.
Example:
– A planet with a higher mass will have a higher escape velocity.
– A smaller planet or a planet with lower mass will have a lower escape velocity.
In summary:
– Escape velocity is directly proportional to the square root of the mass of the planet.
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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What is the gravitational force at the center of Earth?
At the center of Earth, gravitational forces from all sides cancel out and result in zero force. This question is 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/nceRead more
At the center of Earth, gravitational forces from all sides cancel out and result in zero force. This question is 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/
If kinetic energy of a body is increased by 300% then percentage change in momentum will be
To calculate the percentage change in momentum when the kinetic energy increases by 300%, let’s use the relationship between kinetic energy and momentum. Step 1: Relationship between kinetic energy and momentum Kinetic energy (K.E.) is given by: K.E. = (1/2) m v² Momentum (p) is given by: p = m v WeRead more
To calculate the percentage change in momentum when the kinetic energy increases by 300%, let’s use the relationship between kinetic energy and momentum.
Step 1: Relationship between kinetic energy and momentum
Kinetic energy (K.E.) is given by:
K.E. = (1/2) m v²
Momentum (p) is given by:
p = m v
We know:
K.E. = p² / (2m)
Step 2: Express the relationship between K.E. and p
Rearranging:
p² = 2m K.E.
Step 3: Analyze the change in K.E.
If kinetic energy increases by 300%, then the new K.E. is:
New K.E. = Initial K.E. × (1 + 300/100)
New K.E. = 4 × Initial K.E.
.
Step 4: Calculate the change in momentum
From the relationship (p² ∝ K.E.):
If K.E. increases by a factor of 4, then:
New p = √(4 × Initial p²)
New p = 2 × Initial p
Thus, momentum increases by 100%.
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