Problem 744 - JEE Physics

Question

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

Step-by-Step Solution

Answer

The escape velocity on the Moon's surface is approximately 0.84 kms⁻¹, which is closest to option (B) \(0.5\) kms⁻¹.

Step 1: Write down the given values for mass and radius ratios of Moon

Mass of Moon is \(\frac{1}{718}\) times the mass of Earth (Mₑ): Mₘ = \(\frac{1}{718} Mₑ\) Radius of Moon is \(\frac{1}{4}\) times the radius of Earth (Rₑ): Rₘ = \(\frac{1}{4} Rₑ\) Escape velocity on Earth's surface is: Vₑ = 11.2 kms⁻¹ We need to find escape velocity on Moon's surface (Vₘ).

Step 2: Write down the formula for escape velocity and substitute values for Earth

Escape Velocity = \(\sqrt{\frac{2GM}{R}}\) For escape velocity on Earth's surface (Vₑ): Vₑ = \(\sqrt{\frac{2GMₑ}{Rₑ}}\)

Step 3: Divide escape velocities of Moon and Earth

Divide escape velocities of Moon (Vₘ) and Earth (Vₑ) to find the ratio: \(\frac{Vₘ}{Vₑ}\) = \(\frac{\sqrt{\frac{2G(\frac{1}{718} Mₑ)}{\frac{1}{4} Rₑ}}}{\sqrt{\frac{2GMₑ}{Rₑ}}}\)

Step 4: Simplify the equation and plug in the known values

The gravitational constant (G) and Earth's mass (Mₑ) cancel each other out, simplifying the equation: \(\frac{Vₘ}{Vₑ}\) = \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\) Given that the escape velocity on Earth's surface is Vₑ = 11.2 kms⁻¹, plug in the value and solve for Vₘ: Vₘ = Vₑ × \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\) = 11.2 × \(\sqrt{\frac{\frac{1}{718}}{\frac{1}{4}}}\)

Step 5: Calculate the escape velocity on Moon and choose the correct option

Now we can find the escape velocity on Moon's surface: Vₘ = 11.2 × \(\sqrt{\frac{4}{718}}\) = 11.2 × \(\sqrt{\frac{1}{179}}\) Vₘ ≈ 0.84 kms⁻¹ From the given options, none of them exactly match 0.84 kms⁻¹. However, the closest option is (B) \(0.5\) kms⁻¹, which is the most accurate option among the given choices.

Key Concepts

Gravitational Constant

The gravitational constant, often symbolized by the letter \( G \), is an essential concept when discussing gravitational forces and movements. It is a fundamental constant that appears in Newton's law of universal gravitation, which relates the gravitational force between two objects to their masses and the distance between them. The equation for gravitational force is given by:\[ F = \frac{G \times M_1 \times M_2}{r^2} \]In this formula, \( M_1 \) and \( M_2 \) are the masses of the two objects, \( r \) is the distance between the centers of the two objects, and \( G \) is the gravitational constant. This constant helps us understand how gravity functions across different celestial bodies.

• Gravitational force between two bodies is directly proportional to the product of their masses.
• Gravitational force is inversely proportional to the square of the distance between the centers of the two bodies.
• The gravitational constant \( G \) is a fundamental value used to describe the strength of gravity in a universal context.

Understanding \( G \) is crucial when calculating escape velocity, as it affects how fast an object must travel to break free from a celestial body's gravitational pull. This principle significantly impacts calculations for planets, stars, and moons, among other celestial entities.

Mass and Radius Ratios

Mass and radius ratios are crucial when calculating escape velocities from various celestial bodies, such as the Moon and the Earth.These ratios tell us how the attributes of one body compare to another, impacting their gravitational pulls and escape velocities.Let's break this down:

• Mass ratio: This compares the mass of one body to another. Here, the Moon's mass is \( \frac{1}{718} \) times the mass of Earth.
• Radius ratio: This compares the radius of one body to another. The Moon's radius is \( \frac{1}{4} \) that of the Earth.

These ratios help simplify more complex equations by allowing us to substitute these comparable values into calculations. For instance, in the formula for escape velocity \( V = \sqrt{\frac{2GM}{R}} \), the mass and radius values drastically influence the speed required to escape a body's gravitational pull.Using the ratios:- We can adjust the escape velocity formula for the Moon based on its size and mass in relation to Earth.- Calculate how these ratios affect the speed needed for a spacecraft to leave the surface of the Moon.By understanding these ratios, students can better comprehend how differing sizes and masses of celestial bodies impact space travel.

Moon vs Earth Comparison

Comparing the Moon and Earth provides insights into why certain physical phenomena occur differently on each body. The difference in their masses and radii leads to drastic differences in their gravitational fields and subsequently their escape velocities.**Differences in Mass and Radius:**

• The Earth is significantly more massive than the Moon, with the Moon's mass being only \( \frac{1}{718} \)th of Earth's mass.
• The Moon's radius is much smaller, being just \( \frac{1}{4} \) the radius of Earth.

**Impact on Escape Velocity:**- On Earth, the escape velocity is \( 11.2 \) km/s, which is significantly higher than that on the Moon.- With a lower mass and smaller size, the Moon's gravitational pull is weaker, rendering a lower escape velocity approximately \( 0.84 \) km/s.- This means less energy is required to leave the Moon's gravitational field compared to Earth's.Understanding these comparisons helps illustrate why Apollo missions were able to land on and return from the Moon more easily than if they had attempted to reach another body like Earth. These variations provide valuable insights into how we explore and utilize celestial mechanics and gravitation for space travel.