Question

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is \(3 \times 10^{5}\) times heavier than the Earth and is at a distance \(2.5 \times 10^{4}\) times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is \(v_{e}=11.2 \mathrm{~km} \mathrm{~s}^{-1} .\) The minimum initial velocity \(\left(v_{S}\right)\) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) [A] \(v_{S}=22 \mathrm{~km} \mathrm{~s}^{-1}\) [B] \(v_{S}=42 \mathrm{~km} \mathrm{~s}^{-1}\) [C] \(v_{S}=62 \mathrm{~km} \mathrm{~s}^{-1}\) [D] \(v_{S}=72 \mathrm{~km} \mathrm{~s}^{-1}\)

Short answer

The initial velocity required is closest to [B] \(v_{S}=42 \mathrm{~km} \mathrm{~s}^{-1}\).

Step-by-step solution

Understand the concept of escape velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a massive body without further propulsion. This problem involves escaping the gravitational fields of both the Earth and the Sun.

Calculate escape velocity from the Earth

The escape velocity from Earth's gravitational field is given by the problem as \(v_e = 11.2 \text{ km s}^{-1}\). We will use this value directly for the calculations.

Calculate the escape velocity from the Solar System

To find the escape velocity from the Sun (which the rocket must also achieve), we use the formula \(v_s = \sqrt{2GM/R}\), where G is the gravitational constant, M is the mass of the Sun, and R is the distance from the Sun to the Earth. However, since the problem provides relative measures, we can relate Earth's escape velocity to the Sun's escape velocity. Let \(v_S\) denote the escape velocity from the Sun. Because the Sun's mass is \(3 \times 10^5\) times that of Earth and its radius is \(2.5 \times 10^4\) times that of Earth's orbit radius, we have \(v_S = v_e \sqrt{(3 \times 10^5)/(2.5\times 10^4)}\).

Solve for the initial velocity \(v_S\)

Plug the values into the escape velocity relation to find \(v_S\). \(v_S = 11.2 \sqrt{(3 \times 10^5)/(2.5\times 10^4)} = 11.2 \sqrt{12} = 11.2 \times 3.46 = 38.752 \text{ km s}^{-1}\).

Key concepts

Gravitational Field

Imagine space as a fabric that bends around any object with mass, such as planets, moons, or stars. This bending creates what we call a gravitational field, an invisible force that pulls objects towards the center of mass. The Earth's gravitational field, for instance, is what keeps us grounded and dictates the escape velocity necessary for rockets to leave its influence.

In physics, we quantify this gravitational attraction with Newton's law of universal gravitation, which states that every point mass attracts every other point mass by a force acting along the line intersecting both points. This force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers: \[ F = G \frac{{m_1 m_2}}{{r^2}} \]
where \(F\) is the force due to gravity, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses, and \(r\) is the distance between the centers of the two masses. This law helps us understand why larger planets have higher escape velocities and how they interact within the solar system.

Rocket Propulsion

Rocket propulsion is a fascinating application of Newton's third law of motion, which states, 'For every action, there is an equal and opposite reaction.' When a rocket engine expels its fuel downwards as exhaust at high speed, the rocket itself is propelled in the opposite direction with an equal force. This is how rockets are able to ascend through the Earth's atmosphere and, if powerful enough, reach outer space.

The effectiveness of rocket propulsion is measured by its ability to change the rocket's velocity, a term we call 'delta-v.' The rocket needs an immense amount of delta-v to overcome Earth's gravitational pull. Engineers achieve this through staging, where different sections of the rocket contain separate engines and fuel. Once the fuel in a stage is expended, that section is detached, making the rocket lighter and more efficient as it ascends. The final stage must provide enough delta-v to reach the required escape velocity. In theory, \[ \text{Delta-v} = v_e + v_{\text{extra}} \]
where \(v_e\) is the escape velocity and \(v_{\text{extra}}\) is any additional speed needed for the mission, such as entering an orbit or compensating for atmospheric drag.

JEE Advanced Physics

The Joint Entrance Examination (JEE) Advanced assesses students aiming for admission into the prestigious Indian Institutes of Technology (IITs). Physics is a critical component of this examination. It challenges students on various topics, one of them being mechanics, which includes understanding gravitational fields, motion, forces, and energy — concepts all central to the study of escape velocity.

For generating solutions to physics problems for JEE Advanced, students must not only grasp the underlying principles such as universal gravitation and rocket propulsion but also apply mathematical formulas and critical thinking to solve complex problems. Problems such as determining the escape velocity involves combining these concepts with calculated precision, as seen in the given example exercise involving the escape velocity from the Earth and the Sun. Therefore, for JEE Advanced preparation, comprehensive understanding and practice in applying these principles to novel scenarios are essential.

Solar System Dynamics

The motions of celestial bodies within the solar system - including the planets, moons, asteroids, and comets - are all dictated by the gravitational interactions among them and the Sun. Solar system dynamics is a branch of astrophysics that studies these interactions and predicts the movements of these bodies using the laws of physics, especially Newton's law of gravitation.

The concept of escape velocity relates to solar system dynamics in that it defines the velocity an object must reach to break free from the gravitational influence of a celestial body. While studying solar system dynamics, understanding escape velocities is crucial as it not only pertains to artificial objects, like spacecraft, but can also apply to natural phenomena, like the ejection of comets from the solar system or moon-forming processes. The combined gravitational forces exerted by the Sun and the planets can lead to complex orbital motions and even ejections from the solar system, highlighting the delicate balance of gravitational forces at play.