Problem 82 - Master Resource Book in JEE Main Physics

Question

The escape velocity from the earth is \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth would be (a) \(5.5 \mathrm{kms}^{-1}\) (b) \(11 \mathrm{kms}^{-1}\) (c) \(15.5 \mathrm{kms}^{-1}\) (d) \(22 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity from the new planet is \(22 \mathrm{kms}^{-1}\) (Option d).

Step 1: Understand the Escape Velocity Formula

The escape velocity from a planet is given by the formula \( v = \sqrt{\frac{2 G M}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. Since the mean density \( \rho = \frac{M}{V} \), where \( V \) is the volume, and volume of a sphere \( V = \frac{4}{3} \pi R^3 \), the mass \( M = \rho \times \frac{4}{3} \pi R^3 \). This gives the escape velocity formula as \( v = \sqrt{\frac{2 G \rho \cdot \frac{4}{3} \pi R^3}{R}} \), or simplified as \( v = \sqrt{\frac{8 \pi G \rho R^2}{3}} \).

Step 2: Set Expressions for Earth and the New Planet

For Earth, we have \( v_e = 11 \mathrm{kms}^{-1} = \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \). For the new planet (with twice the radius as Earth and the same density), the escape velocity \( v_p = \sqrt{\frac{8 \pi G \rho_e (2R_e)^2}{3}} \).

Step 3: Simplify New Planet's Escape Velocity

Since \( R_p = 2R_e \) and \( \rho_p = \rho_e \), substituting in the new planet's escape velocity gives \( v_p = \sqrt{\frac{8 \pi G \rho_e (2R_e)^2}{3}} = \sqrt{\frac{8 \pi G \rho_e \cdot 4R_e^2}{3}} = 2 \times \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \).

Step 4: Calculate and Compare the Results

Since \( \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} = v_e \), we have \( v_p = 2v_e = 2 \times 11 \mathrm{kms}^{-1} = 22 \mathrm{kms}^{-1} \). Thus, the escape velocity for the new planet is \( 22 \mathrm{kms}^{-1} \).

Key Concepts

Gravitational Constant

Understanding the gravitational constant is crucial in the study of celestial mechanics. Denoted by the symbol \( G \), the gravitational constant is a key value that helps measure the force of attraction between two bodies due to gravity.
It plays an essential role in determining how objects move in a gravitational field. - **Value of \( G \):** Typically, \( G \) has a value of approximately \( 6.674 \, \times \, 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). This very small value highlights how weak the gravitational force is compared to other fundamental forces. - **Importance in Formulas:** In the formula for escape velocity \( v = \sqrt{\frac{2GM}{R}} \), \( G \) allows us to calculate the speed needed for an object to break free from a planet’s gravitational pull. - **Effect on Planetary Motion:** The gravitational constant is crucial for calculating the potential energy and movement of objects in orbits around larger bodies, like planets around stars or moons around planets.By understanding \( G \), students can better grasp how planets, moons, and other celestial bodies interact through the fundamental force of gravity.

Planetary Density

Planetary density is a measure of how much mass is contained within a planet’s volume.
It is intimately connected with the escape velocity and affects how strongly a planet's gravity pulls on nearby objects. - **Definition:** The density \( \rho \) of a planet is given by the formula \( \rho = \frac{M}{V} \), where \( M \) is the mass and \( V \) is the volume. - **Relation with Volume:** Since planets are mostly spherical, their volume \( V \) is calculated using the formula \( \frac{4}{3} \pi R^3 \). Therefore, density is indirectly connected to the radius through volume. - **Impact on Escape Velocity:** In the formula \( v = \sqrt{\frac{8 \pi G \rho R^2}{3}} \), density directly influences the required escape velocity. The denser a planet, the higher the escape velocity required because mass is a component of gravitational attraction.By understanding density, one can predict how easily or difficult it is for objects to reach escape velocity and leave a planet's gravitational influence.

Radius of a Planet

The radius of a planet is a crucial factor that influences numerous other characteristics, such as gravitational pull and escape velocity.
It can drastically alter the conditions on a planet’s surface. - **Definition and Measurement:** A planet's radius is the distance from its center to its surface. This measurement is essential in various calculations involving gravitational force. - **Connection with Escape Velocity:** In the escape velocity formula \( v = \sqrt{\frac{2GM}{R}} \), the radius \( R \) plays a fundamental role. A larger radius requires a higher escape velocity assuming density remains constant. - **Influence on Surface Gravity:** Gravity at a planet's surface is also determined by its radius and mass. A larger planet with the same mass as a smaller one will have weaker surface gravity.Understanding the radius helps in predicting physical and potentially even life-supporting conditions on different planets, making it a fundamental parameter in the field of astronomy.