Problem 761 - JEE Physics

Question

If \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are represent the escape velocity and orbital velocity of satellite corresponding to a circular orbit of -adius \(\mathrm{r}\), then A) \(\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{o}}\) (B) \(\sqrt{2} \mathrm{~V}_{\mathrm{o}}=\mathrm{V}_{\mathrm{e}}\) C) \(\mathrm{V}_{\mathrm{e}}=\left(\mathrm{V}_{\mathrm{O}} / \sqrt{2}\right)\) (D) \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are not related

Step-by-Step Solution

Answer

The short answer is: \(V_e = \sqrt{2}V_o\). Option (B) is correct.

Step 1: 1. Find the formula for escape velocity

The escape velocity is the minimum velocity an object needs to break free from a massive body's gravitational pull. The formula for escape velocity is given by: \(V_e = \sqrt{ \dfrac{2GM}{r} }\) where G is the gravitational constant, M is the mass of the massive body (like Earth), and r is the distance from the center of the massive body (radius of the orbit).

Step 2: 2. Find the formula for orbital velocity

The orbital velocity is the velocity needed for an object to stay in a stable, circular orbit around a massive body. The formula for orbital velocity is given by: \(V_o = \sqrt{ \dfrac{GM}{r} }\) where G is the gravitational constant, M is the mass of the massive body (like Earth), and r is the distance from the center of the massive body (radius of the orbit).

Step 3: 3. Compare escape and orbital velocities

Now let's compare the formulas for escape and orbital velocities: \(V_e = \sqrt{\dfrac{2GM}{r}} \) and \( V_o = \sqrt{\dfrac{GM}{r}} \) To find the relationship between Ve and Vo, we can try to express Ve in terms of Vo. Let's square both Ve and Vo: \(V_e^2 = \dfrac{2GM}{r}\) and \(V_o^2 = \dfrac{GM}{r}\) Now, let's divide Ve^2 by Vo^2: \(\dfrac{V_e^2}{V_o^2} = \dfrac{\dfrac{2GM}{r}}{\dfrac{GM}{r}}\) This simplifies to: \(\dfrac{V_e^2}{V_o^2} = 2\) Now take the square root of both sides: \( \dfrac{V_e}{V_o} = \sqrt{2}\) From this relationship, we can see that: \(V_e = \sqrt{2}V_o\) So, the correct option is (B) \(\sqrt{2}V_o = V_e\).

Key Concepts

Orbital Velocity

Orbital velocity is crucial for understanding how objects move around massive bodies like planets or stars. Essentially, it is the speed needed to maintain a stable orbit. This is not about launching into deep space but staying in a path around the body.
Calculating orbital velocity involves some simple physics. The formula is \( V_o = \sqrt{ \dfrac{GM}{r} } \). Here, \( G \) is the universal gravitational constant, \( M \) is the mass of the massive body, and \( r \) is the radius of the orbit.

• **Governed by Gravity:** It's the balance between the gravitational pull from the massive body and the object's inertia.
• **Stability:** Too fast, and the object shoots away; too slow, and it might crash back down.
• **Applications:** Key for satellites, which rely on precise velocities to ensure they remain in their designated orbital paths.

Gravitational Constant

The gravitational constant, represented as \( G \), is fundamental in physics, especially when discussing celestial mechanics. It's a proportionality factor used in the equation for gravitational force and plays a vital role in calculating both escape and orbital velocities.Known as the symbol that holds the universe together, \( G \) has a value of approximately \( 6.674 \times 10^{-11} \) N(m/kg)^2.

• **Universal Principle:** It implies that all objects exert gravitational pull on each other, a concept discovered by Isaac Newton.
• **Interplanetary Calculations:** Essential for deriving forces and predicting movements in space and astronomy.

The significance of \( G \) extends beyond equations—it connects directly to understanding how planets and stars influence each other across vast distances.

Circular Orbit

A circular orbit may seem simple, but it's a meticulously balanced situation where the gravitational pull of a massive body matches the object's momentum perfectly. It means that as the satellite or object travels around the massive body, it follows a consistent circular path due to this balance.A critical factor here is speed. Circular orbits occur when the object's velocity aligns precisely with the conditions set by the gravitational forces at play. The formula used earlier for orbital velocity \( V_o = \sqrt{ \dfrac{GM}{r} } \) is derived assuming such an orbit.

• **Examples in Our Solar System:** Much like the moon's orbit around Earth and the planets orbiting the Sun.
• **Advantages:** Such orbits are stable and predictable, making them desirable for particular space missions, like communication satellites.

Circular orbits provide a reference standard against which other types of orbits, like elliptical ones, are compared.

Mass of the Massive Body

The mass of the massive body, commonly denoted as \( M \), is a crucial factor in determining the dynamics of orbits and the escape velocities necessary to overcome a body's gravity.Why is it important? Simply put, more mass means more gravitational force. This direct relationship means that the greater the mass of the planet or star, the higher the escape and orbital velocities required.Take Earth, for example:

• **Earth's Influence:** Earth's mass determines the gravitational pull that keeps the moon in orbit and defines the strength of gravity we're all familiar with.
• **Astrophysical Calculations:** Knowing \( M \) is essential for space travel planning, satellite launching, and understanding celestial phenomena.

The mass doesn't just affect nearby objects; it influences celestial mechanics throughout the universe. Each planet, star, or moon’s mass dictates how they interact with others, playing a pivotal role in creating the cosmic ballet of the universe.