Escape velocity depends on mass and radius.
The Gravitation chapter in Class 11 Physics NCERT for CBSE 2024-2025 explores the universal law of gravitation and its practical implications. It covers essential topics such as Kepler’s laws of planetary motion gravitational potential energy acceleration due to gravity and satellite motion. The chapter also includes the concepts of gravitational force escape velocity orbital motion and satellite energy. Multiple-choice questions assess the understanding of variations in gravitational acceleration and the conditions necessary for geostationary orbits. This chapter provides a solid foundation in celestial mechanics and helps students understand the forces governing the motion of objects in space and on Earth.
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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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