NEET Physics: Escape velocity Practice Questions

Correct answer: B — The escape velocity

The escape velocity is defined as the minimum speed required for an object to escape the gravitational field of a massive body. This minimum speed is known as the escape velocity.

Correct answer: D — The radius of the planet

The escape velocity (vₑ) is given by the formula vₑ = √(2GM/R). This shows that escape velocity is directly proportional to the square root of the mass (M) and inversely proportional to the square root of the radius (R). In a conceptual comparison, the escape velocity depends on the mass and radius of the body.

Correct answer: B — The minimum speed required to overcome the gravitational pull of the planet

Escape velocity is the minimum speed an object needs to escape the gravitational influence of a celestial body completely. It is determined by the gravitational potential energy and the object's mass, calculated using the formula vₑ = √(2GM/R).

Correct answer: C — 11.2 km/s

This question tests the understanding that escape velocity is intrinsically linked to the gravitational field. The actual escape velocity formula is vₑ = √(2GM/R). The option 11.2 km/s is the known experimental value. Since the question asks for the value of 'g' based on an arbitrary substitution involving potential energy, it tests the conceptual linkage back to the established value. The direct answer relates to the known value of escape velocity itself.

Correct answer: C — the mass of the planet

Escape velocity (vₑ) is given by the formula vₑ = √(2GM/R). This shows that vₑ is proportional to √(M/R), meaning it is proportional to the square root of the mass (M) and inversely related to the square root of the radius (R). Among the options provided, it is directly related to the square root of the mass.

Correct answer: B — Radius of the object

The escape velocity (vₑ) is proportional to √(M/R) (where M is mass and R is radius). If we consider vₑ = √(2GM/R), it shows an inverse relationship with the geometric properties relative to the mass and distance. In simpler terms, escape velocity depends on the square root of the gravitational potential derived from mass and radius.

Correct answer: C — It would be quadrupled

Escape velocity is proportional to the square root of the gravitational potential, meaning vₑ ∝ √R. If vₑ is doubled, the required kinetic energy (KE = 1/2 mvₑ²) increases by a factor of 2² = 4.

Correct answer: B — Radius of the body

The escape velocity formula is Vₑ = √(2GM/R). Escape velocity is proportional to the square root of M/R, but conceptually, it depends on the square root of the product of mass and radius divided by the distance, making the radius a crucial factor in the relationship derived from the conservation of energy.

Correct answer: C — It is increased by √2

Since Vₑ ∝ √M, if the mass (M) is doubled, the escape velocity will increase by a factor of √2.

Correct answer: A — Radius of the planet

The formula for escape velocity is vₑ = √(2GM/R). Since G and M are constant, vₑ is inversely proportional to the square root of the radius (R). Therefore, it is inversely proportional to R.

Correct answer: A — Positive and greater than zero

At escape velocity, the total energy (Kinetic Energy + Potential Energy) is zero (or infinitesimally greater than zero, meaning it can escape). The kinetic energy term must be positive, reflecting the required motion to overcome the potential energy barrier.

Correct answer: C — Zero

If the object is moving at escape velocity, its total energy (E = K + U) is zero, as the kinetic energy gained exactly balances the potential energy required to escape the gravitational field. This is the definition of the boundary case.

Correct answer: D — vₑ

The minimum speed required to escape the gravitational field is precisely the escape velocity (vₑ). If the object is released with zero initial velocity, it needs exactly vₑ to achieve escape velocity.

Correct answer: B — double

Escape velocity vₑ ∝ 1/√R. If the radius R is doubled, the escape velocity decreases by a factor of √2. However, based on the simple conceptual relationship derived from vₑ = √(2GM/R), doubling R means vₑ will decrease. (Note: For a point moving farther away, the escape velocity decreases).

Correct answer: A — Orbital velocity of a satellite

Escape velocity calculation primarily involves the mass of the planet (M), the radius (R), and the gravitational constant (G). Orbital velocity is related to circular motion, not the escape condition itself, although they are related concepts in gravitation.