A particle is projected vertically upwards from the surface of Earth (radius R) with a kinetic energ

Q: Why can't we use the standard kinematic equation $v^2 = u^2 - 2gh$ to solve this problem?

The equation $v^2 = u^2 - 2gh$ assumes that the acceleration due to gravity $g$ is constant. However, as the particle rises to heights comparable to the Earth's radius $R$, the gravitational force decreases significantly ($g \propto 1/r^2$). Therefore, we must use gravitational potential energy instead of assuming a constant $g$.

Q: Why is the gravitational potential energy negative?

Gravitational potential energy is defined as zero at an infinite distance from the Earth. Since gravity is an attractive force, work must be done against gravity to move a particle to infinity. Therefore, the potential energy at any finite distance is lower than zero, making it negative.

Q: What is the difference between escape velocity and escape energy?

Escape velocity ($v_{esc} = \sqrt{\frac{2GM}{R}}$) is the minimum speed required for a body to escape a gravitational field, whereas escape energy ($E_{esc} = \frac{GMm}{R}$) is the minimum kinetic energy required. Escape energy depends on the mass of the projected particle, while escape velocity is independent of it.

Q: What happens if the initial kinetic energy is greater than the escape energy?

If the initial kinetic energy is greater than the escape energy, the particle will escape Earth's gravity completely and continue to move through space with a non-zero terminal speed (residual kinetic energy) even at an infinite distance.

Q: Does the angle of projection affect the maximum height reached?

No, as long as the particle does not hit the Earth's surface, the maximum height reached depends only on the initial mechanical energy, because gravity is a conservative central force. However, vertical projection ensures the maximum radial distance from the surface.

Question

A particle is projected vertically upwards from the surface of Earth (radius $R$) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of Earth is:

Sigma Percentile · Accepted Answer

### Introduction to Gravitational Escape

Imagine standing on the surface of the Earth, holding a small stone. If you toss it gently upwards, it rises a few meters, slows down under the relentless pull of gravity, and falls back into your hand.

What if you throw it harder? It goes higher, but the result is inevitably the same.

But what if you could launch it with such immense energy that Earth's gravity could never pull it back? This threshold energy is known as the **escape energy**, and the corresponding speed is the famous **escape velocity** ($v_{esc} \approx 11.2	ext{ km/s}$).

In this problem, we explore a fascinating middle ground: what happens when we give a particle exactly **half** of the energy required to escape? How high will it soar before gravity finally halts its upward journey?

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### Analyzing the Setup

Let us model the Earth as a uniform sphere of mass $M$ and radius $R$. A particle of mass $m$ is projected vertically upwards from the surface of the Earth.

First, let's write down the expression for the **escape energy** ($E_{esc}$). The escape energy is the minimum kinetic energy required to take a particle from the surface of the Earth to infinity, where both its kinetic energy and gravitational potential energy become zero.

Using the principle of conservation of energy, we have:

$$E_{esc} = \frac{GMm}{R}$$

According to the problem, the initial kinetic energy ($K_i$) given to the particle is exactly half of this escape energy:

$$K_i = \frac{1}{2} E_{esc} = \frac{GMm}{2R}$$

Since this energy is strictly less than the escape energy, the particle cannot escape to infinity. It will reach a maximum height $h$ above the surface of the Earth, where it will momentarily come to rest before falling back down.

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### The Master Equation: Conservation of Mechanical Energy

Since gravity is a conservative force, the total mechanical energy of the system (kinetic energy + potential energy) remains conserved throughout the motion.

Let us define our initial state (at the surface of the Earth) and final state (at the maximum height $h$):

1. **Initial State (at the surface, distance $r = R$):**
   - Kinetic Energy: $K_i = \frac{GMm}{2R}$
   - Gravitational Potential Energy: $U_i = -\frac{GMm}{R}$

2. **Final State (at maximum height $h$, distance $r = R + h$):**
   - Kinetic Energy: $K_f = 0$ (since the particle momentarily stops at the highest point)
   - Gravitational Potential Energy: $U_f = -\frac{GMm}{R+h}$

Applying the **Conservation of Mechanical Energy**:

$$K_i + U_i = K_f + U_f$$

Substituting our expressions into this master equation:

$$\frac{GMm}{2R} - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$$

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### Step-by-Step Algebraic Simplification

Let us simplify this equation. Notice that the term $GMm$ is common to all terms on both sides. Dividing the entire equation by $GMm$ yields:

$$\frac{1}{2R} - \frac{1}{R} = -\frac{1}{R+h}$$

Now, let's combine the terms on the left-hand side:

$$-\frac{1}{2R} = -\frac{1}{R+h}$$

Multiplying both sides by $-1$:

$$\frac{1}{2R} = \frac{1}{R+h}$$

By taking the reciprocal of both sides, we get:

$$2R = R + h$$

Subtracting $R$ from both sides, we find the final elegant result:

$$h = R$$

This means that when a particle is projected with half of the escape energy, it rises to a height **exactly equal to the radius of the Earth** above the surface!

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### Generalization: The Fractional Energy Formula

What if we want to solve this for any fraction of the escape energy? Let the initial kinetic energy be a fraction $\eta$ of the escape energy:

$$K_i = \eta E_{esc} = \eta \frac{GMm}{R} \quad (	ext{where } 0 < \eta < 1)$$

Using the same energy conservation approach:

$$\eta \frac{GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$$

Dividing by $GMm$:

$$\frac{\eta - 1}{R} = -\frac{1}{R+h}$$

$$\frac{1 - \eta}{R} = \frac{1}{R+h}$$

Taking the reciprocal:

$$R + h = \frac{R}{1 - \eta}$$

$$h = R \left( \frac{1}{1 - \eta} - 1 ight)$$

$$h = \left( \frac{\eta}{1 - \eta} ight) R$$

This is a highly powerful generalized formula! Let's test it:
- For $\eta = 1/2$ (our problem): $h = \frac{1/2}{1 - 1/2} R = R$. (Matches perfectly!)
- For $\eta = 3/4$: $h = \frac{3/4}{1 - 3/4} R = 3R$.

This elegant relationship shows how physics beautifully connects simple algebraic fractions to cosmic scales.

Animated Solution for Physics - Gravitation: A particle is projected vertically upwards from the surface of Earth (radius $R$ ) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of Earth is:

Visualized Solution

Visualizing the Setup

Understanding Escape Energy

Initial Kinetic Energy Given

Conservation of Mechanical Energy

Gravitational Potential Energy Setup

Energy at Maximum Height

Substituting Values into the Master Equation

Simplifying the Equation

Solving for Height $h$

The Way Forward

The Sigma Insight: Gravitational Potential and Potential Energy

Introduction to Gravitational Escape

Analyzing the Setup

The Master Equation

Step-by-Step Algebraic Simplification

Generalization

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