Problem 746 - JEE Physics

Question

Escape velocity on the surface of earth is \(11.2 \mathrm{kms}^{-1}\) Escape velocity from a planet whose masses the same as that of earth and radius \(1 / 4\) that of earth is \(=\ldots \ldots \ldots \mathrm{kms}^{-1}\) (A) \(2.8\) (B) \(15.6\) (C) \(22.4\) (D) \(44.8\)

Step-by-Step Solution

Answer

The escape velocity of the planet with the same mass as Earth and radius 1/4 that of Earth is \(22.4 \mathrm{kms}^{-1}\) (option C).

Step 1: Write down the given information

The given information: Escape velocity of Earth (\(v_{Earth}\)) = 11.2 km/s Mass of the planet (\(M\)) = Mass of Earth Radius of the planet (\(R\)) = 1/4 * Radius of Earth

Step 2: Write down the formula for escape velocity

Escape velocity formula: \[v = \sqrt{\frac{2GM}{R}}\]

Step 3: Set up proportion for the planet and Earth

We will set up a proportion to relate the escape velocities of the planet and Earth: \[\frac{v_{planet}}{v_{Earth}} = \sqrt{\frac{R_{Earth}}{R_{planet}}}\]

Step 4: Substitute the given values

Substitute the given values into the proportion: \[\frac{v_{planet}}{11.2} = \sqrt{\frac{4R_{planet}}{R_{planet}}}\]

Step 5: Solve for the escape velocity of the planet

Solve the proportion to find the escape velocity (\(v_{planet}\)): \[v_{planet} = 11.2 * \sqrt{4}\] \[v_{planet} = 11.2 * 2 = 22.4 \mathrm{kms}^{-1}\] So, the escape velocity of the planet is 22.4 km/s, which corresponds to option C.

Key Concepts

Gravitational Constant

The gravitational constant, denoted as \( G \), is a crucial figure in physics that helps us understand how gravity works on a universal scale. It appears in Newton's law of universal gravitation, which describes the attractive force between two masses. This force is essential in keeping planets in their orbits and causes objects to fall towards the ground on Earth.

The formula for gravitational force is:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]where:

• \( F \) is the gravitational force between the two masses
• \( m_1 \) and \( m_2 \) are the two masses
• \( r \) is the distance between the centers of their masses
• \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)

In the context of escape velocity, the gravitational constant helps us calculate how fast an object needs to go to break free from a planet's gravitational pull. It's a constant that applies everywhere in the universe, making it fundamental in both calculating escape velocity and modeling gravitational interactions between celestial bodies.

Planetary Mass

Planetary mass refers to the total amount of matter contained in a planet. It is a significant factor in determining the gravitational pull a planet exerts on objects. A planet's mass, along with its radius, affects the escape velocity of an object from its surface.

Escape velocity is influenced by mass according to the formula:\[v = \sqrt{\frac{2GM}{R}}\]Here, \( M \) is the mass of the planet, and \( G \) and \( R \) are the gravitational constant and radius of the planet, respectively. As we can see from this equation, a greater mass increases the escape velocity. This means that for a more massive planet, more energy is required for an object to escape its gravitational field.

In the given exercise, the mass of the new planet is the same as Earth's. This simplifies our calculations because we can directly compare escape velocities based on changes in planetary radius without recalculating based on a different mass.

Planetary Radius

The radius of a planet is measured from its center to its surface and plays a critical role in determining escape velocity. The formula for escape velocity shows that as the radius decreases, the velocity required for escape increases:

\[v = \sqrt{\frac{2GM}{R}}\]This equation indicates that the relationship between escape velocity and radius is inverse: as radius halved, for example, the escape velocity increases.

In our exercise example, the radius of the new planet is \( \frac{1}{4} \) of Earth's radius. This means that the gravitational pull at the surface is more concentrated, which leads to a higher escape velocity. This inverse relationship helps explain why less radius requires more speed to break free from the gravitational influence of a planet.

• Smaller radius results in higher escape velocity.
• This is why compact celestial bodies often have significant gravitational fields compared to their size.

Understanding how radius affects gravity allows us to predict the behavior of objects, such as satellites or spacecraft, around different planets or celestial bodies.