Problem 739 - JEE Physics

Question

The escape velocity of a body from earth's surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth's surface will be (A) \((\mathrm{Ve} / \sqrt{2})\) (B) \((\mathrm{Ve} / 2)\) (C) \((\mathrm{Ve} / 2 \sqrt{2})\) (D) \((\mathrm{Ve} / 4)\)

Step-by-Step Solution

Answer

The escape velocity of the same body from a height equal to 7R from Earth's surface is \(V_{e1} = \frac{V_e}{2}\). The correct answer is option (B).

Step 1: Write the formula for escape velocity from Earth's surface.

Using the formula for escape velocity: \(V_e = \sqrt{\frac{2GM}{r_e}}\) Here, the mass of Earth is M and the distance from the Earth's center to its surface (Earth's radius) is R.

Step 2: Write the formula for escape velocity at a height of 7R from Earth's surface.

Now, we will write the formula for escape velocity at a height of 7R from the Earth's surface. Let the new escape velocity be \(V_{e1}\). The distance from the Earth's center at this height becomes r = 8R. \(V_{e1} = \sqrt{\frac{2GM}{8R}}\)

Step 3: Compare the two escape velocities.

Now we need to find the relation between \(V_e\) and \(V_{e1}\). To do this, we will divide the equations of \(V_{e1}\) by \(V_e\). \(\frac{V_{e1}}{V_e} = \frac{\sqrt{\frac{2GM}{8R}}}{\sqrt{\frac{2GM}{R}}}\)

Step 4: Simplify the equation to find the ratio between the two escape velocities.

Simplify the equation to find the ratio between \(V_{e1}\) and \(V_e\): \(\frac{V_{e1}}{V_e} = \frac{\sqrt{\frac{1}{4}}}{\sqrt{1}} = \frac{1}{2}\) So, \(V_{e1} = \frac{V_e}{2}\). Therefore, the escape velocity of the same body from a height equal to 7R from Earth's surface is: \(V_{e1} = \frac{V_e}{2}\) The correct answer is option (B).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. This energy depends on the mass of the object, the height of the object from a reference point (usually the surface of the Earth), and the gravitational force acting on it. Mathematically, it can be expressed as:
\[ U = mgh \] where:

• \(U\) is the gravitational potential energy.
• \(m\) is the mass of the object.
• \(g\) is the acceleration due to gravity.
• \(h\) is the height above the reference point.

When discussing larger distances from the Earth, such as the escape velocity scenario, we use a more universal formula that accounts for these distances:
\[ U = -\frac{GMm}{r} \] Here, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the Earth's center. This formula showcases that the farther you are from the Earth, the higher the potential energy, which means less energy is needed to "escape" the Earth's gravity.

Orbital Mechanics

Orbital mechanics is a crucial part of understanding how bodies move through space under the influences of mutual gravitational attraction. A key concept in orbital mechanics is escape velocity, which is the minimum speed an object needs to reach in order to break free from the gravitational pull of a celestial body, like Earth, without any further propulsion.
This speed depends on the mass and radius of the body in question. For Earth, this is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{r_e}} \] where:

• \(V_e\) is the escape velocity.
• \(G\) is the universal gravitational constant.
• \(M\) is the mass of the Earth.
• \(r_e\) is the radius of the Earth.

As you move to a greater height from Earth's surface, the required escape velocity changes because the distance from the Earth's center increases, reducing the gravitational attraction. In the exercise, at a distance of \(8R\) (since \(7R\) above Earth's surface), the escape velocity used the formula \( V_{e1} = \sqrt{\frac{2GM}{8R}} \), showing how location affects orbital mechanics.

Physics Problem Solving

Solving physics problems effectively requires a structured approach which includes understanding the problem, using known formulas, and carefully solving step-by-step. Let's break down the steps used in the exercise above:

• Identify and Understand: Recognize the key variables and constants in the problem. In this case, identifying the initial and the new scenarios defined by height differences was critical.
• Use the right formulas: Apply the correct physical formulas for the given context. Here, the escape velocity formula was crucial for both the surface and the new height \(7R\) above it.
• Solve Step-by-Step: Divide and simplify the problem components into manageable calculations, making sure each step logically follows from the previous one.
• Compare and Analyze: Lastly, the comparison of results (\(V_{e1}\) and \(V_e\)) was necessary to confirm and verify the solution to the problem, securing that \(V_{e1} = \frac{V_e}{2}\).

These steps often guide through physics problems, breaking them down into simpler tasks ultimately leading to the correct answer.