Physics Q93 – AP-EAMCET 2025

If a body is projected from the surface of the earth with a velocity of

\sqrt{5} V_{E}

, then the velocity of the body when it escapes from the gravitational influence of the earth is (Escape velocity of a body from the surface of the earth,

V_{E} = 11.2 {km s}^{- 1}

)

Clue 1 of 3: Conservation of Mechanical Energy

When a body moves under the influence of a conservative force like gravity, its total mechanical energy (kinetic + potential) remains constant.

🧠 Full Solution Path

Step 1: Identify Initial and Final States and Energies✦ Active

The body is projected from Earth's surface with an initial velocity $v_{i} = \sqrt{5} V_{E}$ . We need to find its final velocity $v_{f}$ when it escapes Earth's gravitational influence (i.e., at an infinite distance from Earth).

Initial Kinetic Energy ( $K E_{i}$ ): $K E_{i} = \frac{1}{2} m v_{i}^{2} = \frac{1}{2} m (\sqrt{5} V_{E})^{2} = \frac{5}{2} m V_{E}^{2}$

Initial Potential Energy ( $P E_{i}$ ): $P E_{i} = - \frac{G M m}{R}$ , where $G$ is the gravitational constant, $M$ is Earth's mass, $m$ is the body's mass, and $R$ is Earth's radius.

Final Kinetic Energy ( $K E_{f}$ ): $K E_{f} = \frac{1}{2} m v_{f}^{2}$

Final Potential Energy ( $P E_{f}$ ): When the body escapes gravitational influence (reaches infinity), its potential energy is zero, so $P E_{f} = 0$ .

💡 Teacher's Secret Hint

Remember that gravitational potential energy is negative and is defined as zero at infinity. On the surface of Earth, it's at its most negative value.

Step 2: Apply Conservation of Mechanical Energy○ Expand

According to the principle of conservation of mechanical energy, the total energy of the body remains constant:

E_{i} = E_{f}

K E_{i} + P E_{i} = K E_{f} + P E_{f}

\frac{5}{2} m V_{E}^{2} - \frac{G M m}{R} = \frac{1}{2} m v_{f}^{2} + 0

💡 Teacher's Secret Hint

Ensure you correctly assign the signs for potential energy. It's negative when attractive.

Step 3: Relate Gravitational Potential Energy to Escape Velocity○ Expand

The escape velocity $V_{E}$ from Earth's surface is defined as:

V_{E} = \sqrt{\frac{2 G M}{R}}

Squaring both sides, we get:

V_{E}^{2} = \frac{2 G M}{R}

From this, we can express the term $\frac{G M}{R}$ as:

\frac{G M}{R} = \frac{V_{E}^{2}}{2}

💡 Teacher's Secret Hint

This relationship is crucial for simplifying the energy conservation equation and solving for the unknown velocity in terms of $V_{E}$ .

Step 4: Substitute and Solve for the Final Velocity○ Expand

Substitute the expression for $\frac{G M}{R}$ into the energy conservation equation from Step 2:

\frac{5}{2} m V_{E}^{2} - m (\frac{V_{E}^{2}}{2}) = \frac{1}{2} m v_{f}^{2}

Divide the entire equation by $m$ (assuming $m \neq 0$ ):

\frac{5}{2} V_{E}^{2} - \frac{1}{2} V_{E}^{2} = \frac{1}{2} v_{f}^{2}

Simplify the left side:

\frac{4}{2} V_{E}^{2} = \frac{1}{2} v_{f}^{2}

2 V_{E}^{2} = \frac{1}{2} v_{f}^{2}

Multiply by 2:

4 V_{E}^{2} = v_{f}^{2}

Take the square root of both sides (considering only the positive velocity):

v_{f} = \sqrt{4 V_{E}^{2}} = 2 V_{E}

💡 Teacher's Secret Hint

Notice how the mass of the body cancels out, indicating that the final velocity is independent of the body's mass.

Step 5: Calculate the Numerical Value○ Expand

Given the escape velocity $V_{E} = 11.2 {km s}^{- 1}$ .

Substitute this value into the expression for $v_{f}$ :

v_{f} = 2 \times 11.2 {km s}^{- 1}

v_{f} = 22.4 {km s}^{- 1}

This matches option 1.

💡 Teacher's Secret Hint

Always double-check your units. Here, the velocity is in km/s, so the final answer should also be in km/s.

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Physics Question 93 – AP-EAMCET 2025

Clue 1 of 3: Conservation of Mechanical Energy

🧠 Full Solution Path