Problem 13 - Master Resource Book in JEE Main Physics

Question

If suppose moon is suddenly stopped and then released (given radius of moon is one-fourth the radius of earth) and the acceleration of moon with respect to earth is \(0.0027 \mathrm{~ms}^{-2}\) ), then the acceleration of the moon just before striking the earth's surface is (Take \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(0.0027 \mathrm{~ms}^{-2}\) (b) \(5.0 \mathrm{~ms}^{-2}\) (c) \(6.4 \mathrm{~ms}^{-2}\) (d) \(10 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

The acceleration is (d) 10 \(\mathrm{ms}^{-2}\).

Step 1: Understand the Problem

The problem states we need to find the acceleration of the moon just before it would theoretically strike the Earth's surface after being stopped and then released. We have, specifically given that the radius of the moon is one-fourth the radius of Earth.

Step 2: Use the Formula for Gravitational Acceleration

The formula for gravitational acceleration due to a planet is given by \[ a = g' = \frac{GM}{r^2} \] where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(r\) is the radius of that planet. Since the moon is stopped and then released, its acceleration just before striking is essentially the acceleration due to gravity at that point.

Step 3: Relate Earth's and Moon's Radius

We know Earth's radius, denoted as \(R_e\), and the moon's radius provided as \(\frac{R_e}{4}\). The acceleration due to gravity at the surface of Earth is provided as \(g = 10 \mathrm{~ms}^{-2}\).

Step 4: Determine Effective Distance to Earth's Center

When the moon is 'just before striking the Earth', it can be considered as being at Earth's surface. Thus, effectively, its total distance from Earth's center is essentially its original distance minus its radius, which is exerted by Earth's gravity, scaled according to distance.

Step 5: Calculate Effective Acceleration

Given the radius relation. If the moon's original gravitational acceleration relative to Earth is \(0.0027 \mathrm{~ms}^{-2}\) far away, when it comes close (exerted by Earth solely), the gravitational acceleration will tend towards Earth's surface gravity as \(g=10 \mathrm{~ms}^{-2}.\)

Step 6: Conclude on the Acceleration

The closer the moon will approach Earth, the higher the effective gravity acting will mimic the acceleration due to Earth's surface gravity, i.e., \(10 \mathrm{~ms}^{-2}\). Hence, the correct answer is closest to Earth surface gravity, meaning (d) \(10 \mathrm{~ms}^{-2}\).

Key Concepts

Acceleration due to gravity

The concept of acceleration due to gravity is central when discussing motion under the influence of gravitational forces. Acceleration due to gravity is essentially the rate at which an object speeds up as it falls freely under the influence of a celestial body's gravity. On Earth, this acceleration is approximately 9.8 m/s², but we commonly simplify it to 10 m/s² for ease of calculation in physics problems.

• The formula used to calculate gravitational acceleration is \[a = \frac{GM}{r^2}\]where \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, and \(r\) is the radius from the center of the body to the object in free fall.
• In the context of the original problem, the moon's acceleration from Earth is reduced to 0.0027 m/s² due to its distance, but as it comes closer, the acceleration increases toward Earth's surface acceleration, 10 m/s².

Understanding this formula helps explain how gravitational acceleration changes based on distance from a mass like Earth.

Gravitational force

Gravitational force is the attraction between two masses, which is explained by Isaac Newton's law of universal gravitation. This universal force acts between any two bodies in the universe that have mass.

• The force is calculated using the formula \[F = \frac{Gm_1m_2}{r^2}\]where \(F\) represents the gravitational force between two objects, \(G\) the gravitational constant, \(m_1\) and \(m_2\) the respective masses of the two objects, and \(r\) the distance separating their centers.
• The force of gravity governs the motion of celestial bodies, such as planets and moons, dictating their orbits and interactions.

This force not only affects massive bodies in space but also applies to every object on Earth, albeit often with negligible impact compared to the planet's own gravity.

Radius and gravitational effect

The radius of a celestial body plays a crucial role in the gravitational effects it exerts on other bodies. The greater the distance from the center of mass, the weaker the gravitational pull.

• Earth's radius is denoted as \(R_e\). In the problem, the moon’s radius is given as one-fourth of Earth’s, impacting the effective gravitational pull when the moon is closer to Earth.
• As distance between objects decreases, gravitational influence increases, making radius a critical factor.
• The gravitational force and acceleration are inversely proportional to the square of distance, implying a small decrease in distance dramatically increases gravitational effects.

On approach, the moon, originally experiencing minimal gravitational pull due to Earth's radius, begins to rapidly feel Earth's gravity when nearing the surface.