Problem 748 - JEE Physics

Question

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be \(\ldots \ldots \ldots \mathrm{kms}^{-1}\) (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

Step-by-Step Solution

Answer

The escape velocity for a body projected at an angle of 45° with the vertical is \(\frac{11}{\sqrt{2}}\) km/s. Therefore, the correct answer is (A) \(\frac{11}{\sqrt{2}}\) km/s.

Step 1: Know the formula of escape velocity

Escape velocity is calculated using the formula: \[v_e = \sqrt{\frac{2GM}{R}}\] where, \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

Step 2: Determine the vertically projected escape velocity

We are given that the escape velocity for a body projected vertically upwards is 11 km/s. Thus, for this case, \(v_e = 11\) km/s.

Step 3: Understand the dependence of escape velocity and projection angle

The escape velocity is directly proportional to the initial velocity along the vertical axis. When the body is projected at an angle of 45°, the vertical component of the initial velocity will be different than when it's projected vertically upwards.

Step 4: Calculate the vertical component of the initial velocity for a 45° projection

When the body is projected at an angle of 45° with the vertical, the vertical component of the initial velocity is given by: \[v_{vertical} = v_e \times \cos{45°}\] As we know that \(\cos{45°} = \frac{1}{\sqrt{2}}\), we can replace it in the equation: \[v_{vertical} = 11 \times \frac{1}{\sqrt{2}}\]

Step 5: Apply the escape velocity formula for the 45° projection

The escape velocity for 45° projection can now be determined using the vertical component of the initial velocity: \[v_e= \sqrt{\frac{2GM}{R}} * v_{vertical}\] Substitute the known values: \[v_e = 11 \times \frac{1}{\sqrt{2}}\] \[v_e = \frac{11}{\sqrt{2}}\, \mathrm{kms}^{-1}\] Thus, the escape velocity for a body projected at an angle of 45° with the vertical is \(\frac{11}{\sqrt{2}}\) km/s. Therefore, the correct answer is (A) \(\frac{11}{\sqrt{2}}\) km/s.

Key Concepts

Gravitational Constant

When we talk about calculating escape velocity, one of the important factors is the gravitational constant, denoted as \(G\). The value of \(G\) is universal, meaning it is the same everywhere in the universe. This constant plays a crucial role in determining the gravitational pull between two bodies. It appears in the formula for escape velocity:

• Escape velocity formula: \(v_e = \sqrt{\frac{2GM}{R}}\)
• Where \(G\) is the gravitational constant.

The gravitational constant \(G\) has a value of approximately \(6.674 \times 10^{-11} \text{Nm}^2\text{/kg}^2\). Understanding this can help us appreciate the immense energy required to overcome the gravitational pull of a planet or celestial body, letting an object break free from its gravitational field. This is why the gravitational constant is a fundamental part of physics and crucial in calculations involving celestial bodies.

Projectile Motion

Projectile motion refers to the motion of an object that is projected into the air and affected only by the force of gravity. In the context of escape velocity, especially at angles, understanding projectile motion helps in calculating the exact velocity needed for the object to overcome Earth's gravity.

• Projectile motion arises due to the combined effects of horizontal and vertical forces.
• It is characterized by a curved path that an object follows after being launched.
• The projection angle plays a critical role in determining how far and at what trajectory the object travels.

In our exercise, a body is projected at a 45° angle, which means that both the vertical and horizontal components of the velocity are equal. The understanding of this concept is critical in breaking down the velocity into components, particularly when working out changes in escape velocity due to varying initial projection angles.

Vertical Component of Velocity

When an object is projected into the air, its initial velocity can be broken down into two components - vertical and horizontal. In our case, the vertical component of velocity is particularly important for escape velocity calculations:

• The vertical component determines how quickly an object moves upwards against gravity.
• For a 45° angle projection, this is calculated as \(v_{vertical} = v_e \times \cos{45°}\).
• For our specific problem, given \(v_e = 11 \text{km/s}\), we find the vertical component as \(\frac{11}{\sqrt{2}} \text{km/s}\).

This critical aspect reflects how efficiently the force is used to counteract gravity. The vertical component is essential for determining how high, and ultimately whether, an object can achieve escape velocity when gravity is considered. Thus, these calculations highlight the necessity of understanding both vertical and horizontal components in projectile motion scenarios.