Problem 752 - JEE Physics

Question

The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(\mathrm{R}=6.96 \times 10^{8} \mathrm{~m}\). The escape velocity of rocket from the sun \(\mathrm{is}=\ldots \ldots \ldots \mathrm{km} / \mathrm{sec}\) \(\begin{array}{llll}\text { (A } 11.2 & \text { (B) } 12.38 & \text { (C) } 59.5 & \text { (D) } 618\end{array}\)

Step-by-Step Solution

Answer

The escape velocity of a rocket from the sun is approximately 617.542 km/sec, which is closest to answer choice D.

Step 1: The formula for escape velocity is given by \(v_{e} = \sqrt{ \frac{2GM}{R} }\), where \(v_{e}\) is the escape velocity, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2}\)), \(M\) is the mass of the celestial body (in this case, the sun), and \(R\) is the radius of the celestial body. #Step 2: Plug in the given values#

We are given the mass of the sun as \(1.99 \times 10^{30} \mathrm{kg}\) and the radius of the sun as \(6.96 \times 10^{8} \mathrm{m}\). We have the gravitational constant \(G = 6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2}\). Plug these values into the escape velocity formula: \(v_{e} = \sqrt{ \frac{2 \times 6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2} \times 1.99 \times 10^{30} \mathrm{kg}}{6.96 \times 10^{8} \mathrm{m}} }\). #Step 3: Calculate the escape velocity in m/sec#

Step 2: Simplify and calculate the escape velocity: \(v_{e} = \sqrt{ \frac{2 \times 6.674 \times 10^{-11} \times 1.99 \times 10^{30}}{6.96 \times 10^{8}} }\) \( \mathrm{m/sec} \approx 617,542 \mathrm{m/sec}\). #Step 4: Convert the escape velocity to km/sec#

To convert the escape velocity from m/sec to km/sec, divide the result by 1000: \(617,542 \mathrm{m/sec} \div 1000 = 617.542 \mathrm{km/sec}\). The escape velocity of a rocket from the sun is approximately 617.542 km/sec, which is closest to answer choice D.

Key Concepts

Gravitational Constant

The gravitational constant, symbolized as \(G\), is a key piece of the puzzle when examining gravitational forces. Its value, \(6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2\), is used in various calculations involving gravity, such as finding the escape velocity, which tells us the speed needed to break free from a celestial body's gravity. The gravitational constant applies universally, meaning it works the same way whether you're dealing with Earth, the sun, or other galaxies. It essentially tells us how strong the gravitational pull is between two masses at a certain distance. For example, when calculating escape velocity, \(G\) helps us gauge the energy needed for an object to leave a planet or star's gravitational field.

• Foundation of Gravity Calculations: \(G\) is the backbone of many formulas in physics.
• Universal Application: It holds true everywhere in the universe, making it a fundamental constant in physics.

Understanding this constant is crucial when learning about topics like escape velocity, where it plays a pivotal role.

Mass of Sun

The mass of the sun is another critical element to consider. It is a colossal \(1.99 \times 10^{30} \, \text{kg}\). This enormous mass is significant because it provides the gravitational force that keeps our solar system together. The sun, being the largest object, dictates the movement of planets and other bodies orbiting around it.In the context of escape velocity, the mass of the sun is directly proportional to it. The more massive a celestial body is, the higher the escape velocity required. This is because the sun’s strong gravitational pull is directly related to its mass.

• Central to Solar System: Provides the necessary gravitation to hold planets within their orbits.
• Influence on Escape Velocity: A greater mass means a higher velocity required to break free from its gravity.

This massive weight highlights the sheer amount of energy a rocket needs to escape the sun’s fierce pull.

Radius of Sun

The radius of the sun, measured at \(6.96 \times 10^{8} \, \text{m}\), is yet another important factor in the equation for escape velocity. The radius connects to how far from the center of mass, we're starting our escape. The larger the radius, the further the gravitational influence stretches, which can affect the calculations of velocity needed for escape. This radius is not just physically impressive; it's a vital part of calculating how gravity diminishes with distance, another reason why escape velocity is sensitive to changes in radius.

• Influences Gravitational Pull: Determines the reach of a celestial body's gravity.
• Plays into Distance: More radius means considering a greater part of the gravitational field when calculating escape velocity.

Knowing the sun’s radius allows us to accurately calculate the speed required to overcome its expansive gravitational force.