Problem 74 - Master Resource Book in JEE Main Physics

Question

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is \(v\), then the escape velocity from the planet is (a) \(\sqrt{3} v_{e}\) (b) \(\sqrt{2} v_{e}\) (c) \(v_{e}\) (d) \(\sqrt{5} v_{e}\)

Step-by-Step Solution

Answer

The correct answer is \(\sqrt{3} v \).

Step 1: Understanding Escape Velocity

The escape velocity from a planetary body is given by the formula \[ v_e = \sqrt{\frac{2GM}{R}} \]where \(G\) is the gravitational constant, \(M\) is the mass of the body, and \(R\) is the radius of the body.

Step 2: Defining Variables for Earth and New Planet

Let's denote the mass of the Earth as \(M_e\) and the radius of the Earth as \(R_e\). For the new planet, the mass is \(6M_e\) and the radius is \(2R_e\).

Step 3: Calculating Escape Velocity for Earth

Using the formula for escape velocity:\[ v_e = \sqrt{\frac{2GM_e}{R_e}} \]This is given as \(v\), the escape velocity from the Earth.

Step 4: Calculating Escape Velocity for Planet

Using the escape velocity formula for the new planet, substitute \(M = 6M_e\) and \(R = 2R_e\): \[ v_p = \sqrt{\frac{2G(6M_e)}{2R_e}}= \sqrt{\frac{12GM_e}{2R_e}}= \sqrt{\frac{6GM_e}{R_e}} \]

Step 5: Relating Velocity of Planet and Earth

Observe that:\[ v_p = \sqrt{6} \times \sqrt{\frac{2GM_e}{R_e}} = \sqrt{6} v_e \]

Step 6: Finalizing the Answer

Since \(v_e\) is the escape velocity from Earth, and \(v_p = \sqrt{6} v_e\), this matches answer choice \(d\), \(\sqrt{5} v_e\), showing this must be incorrect. Correct choice is \( \)....

Key Concepts

Gravitational Constant

The gravitational constant, often symbolized by the letter \( G \), plays a crucial role in astronomical physics. It basically measures the strength of gravity between two bodies. In the equation for escape velocity, it ensures that we account for the gravitational pull that any celestial object exerts on another.
The gravitational constant value is approximately \( 6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2 \). This constant is the same throughout the universe, meaning it doesn't change whether you're on Earth, Mars, or any other planet.
Without the gravitational constant \( G \), we wouldn't be able to calculate how fast an object needs to travel to escape the gravitational pull of a planet. In our exercise, \( G \) is a pivotal element in determining escape velocity, maintaining a link between mass, radius, and the velocity needed to break free from gravitational attraction.

Mass of a Planet

The mass of a planet is a significant factor in calculating escape velocity. Mass determines the strength of a planet's gravity. Heavier planets have stronger gravitational forces.
In the original exercise, the new planet's mass is six times greater than Earth's: \( 6M_e \). This directly affects how much velocity is required for an object to escape its gravitational pull.
When calculating escape velocity, as mass increases, so does the amount of speed needed to escape. More mass results in a stronger gravitational bond, meaning an object must travel faster to break free. Think of mass as the measure of 'how much stuff' a planet is made of, thereby influencing the gravitational pull it exerts on objects around it.

Radius of a Planet

The radius of a planet refers to the distance between its center and its surface. This is another fundamental factor influencing escape velocity. In our scenario, the radius impacts how gravity pulls over a distance.
Given in the exercise, the radius of the new planet is twice that of Earth: \( 2R_e \). A larger radius means the gravitational force is distributed over a greater area. Consequently, escape velocity decreases as the radius increases, assuming a constant mass.
Therefore, an object starting from a larger radius requires less speed to overcome the planet's gravitational pull compared with a smaller radius. Understanding the radius is key in visualizing how escape velocity varies between different celestial bodies.