The initial velocity vi required to project a body vertically upward from the surface of the Earth t

Q: Why is the potential energy negative?

Gravitational potential energy is defined as zero at infinity. Since gravity is an attractive force, work must be done against it to move an object to infinity, meaning the energy closer to Earth is less than zero, hence negative.

Q: Why doesn't the mass of the body matter?

When we equate the initial and final energies, the mass $m$ appears in every single term (both kinetic and potential). Therefore, it cancels out completely. This is a profound feature of gravity—all objects fall (and rise) the same way regardless of their mass!

Q: What if the height was given from the center of the Earth instead of the surface?

If the problem stated the object reached a distance of $10R$ from the center, then our final distance $r$ would just be $10R$. Always read carefully whether the height is from the surface or the center!

Q: How did we get the escape velocity formula?

Escape velocity is the speed needed to reach infinity with zero leftover kinetic energy. Setting $E_{\text{initial}} = E_{\text{final}} = 0$ gives $\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$, which simplifies to $v_e = \sqrt{\frac{2GM}{R}}$.

Q: Can we use kinematic equations like $v^2 = u^2 + 2as$ here?

No! The acceleration due to gravity $g$ is not constant over such large distances. It decreases as you move further from Earth. Kinematic equations only work for constant acceleration, so we must use energy conservation.

Question

The initial velocity $v_i$ required to project a body vertically upward from the surface of the Earth to reach a height of $10R$, where $R$ is the radius of the Earth, may be described in terms of escape velocity $v_e$ such that $v_i = \sqrt{\frac{x}{y}} 	imes v_e$. The value of $x$ will be ............. .
[2021, 25 Feb Shift-II]

Sigma Percentile · Accepted Answer

### The Cosmic Setup

Imagine standing on the surface of the Earth, looking up at the vast expanse of space. You have a ball in your hand, and you want to throw it so hard that it reaches a staggering height of $10R$, where $R$ is the radius of the Earth.

To achieve this monumental feat, you give it an initial velocity of $v_i$. As the ball travels upwards, it fights against the relentless pull of Earth's gravity, slowing down until it reaches its peak and momentarily stops.

### The Master Equation: Energy Conservation

In this cosmic journey, the only force acting on our ball is gravity. Because gravity is a conservative force, we can rely on one of the most powerful principles in physics: the Conservation of Mechanical Energy.

This principle tells us that the total energy of the ball at the moment it leaves your hand must perfectly equal its total energy at the very top of its trajectory.

$$E_{	ext{initial}} = E_{	ext{final}}$$

### Analyzing the Launch

Let's break down the energy at the starting line. The moment the ball is launched, it possesses a kinetic energy due to its speed $v_i$.

$$K_i = \frac{1}{2}mv_i^2$$

Simultaneously, because it is resting on the Earth's surface, it has a gravitational potential energy. Remember, potential energy is negative because it's a bound system!

$$U_i = -\frac{GMm}{R}$$

So, our total initial energy is the sum of these two components.

### Reaching the Peak

Now, let's fast forward to the moment the ball reaches its maximum height of $10R$. At this exact instant, the ball stops moving before falling back down, meaning its kinetic energy is zero.

$$K_f = 0$$

However, its potential energy has changed. The distance is always measured from the center of the Earth. So, the total distance $r$ is the Earth's radius $R$ plus the height $10R$, giving us $11R$.

$$U_f = -\frac{GMm}{11R}$$

### The Grand Equating

Now, we bring our initial and final states together into our master equation.

$$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$$

Our goal is to find the initial velocity $v_i$. Let's move the initial potential energy term to the right side of the equation.

$$\frac{1}{2}mv_i^2 = \frac{GMm}{R} - \frac{GMm}{11R}$$

By taking a common denominator of $11R$, we can easily subtract these fractions.

$$\frac{1}{2}mv_i^2 = \frac{11GMm - GMm}{11R} = \frac{10GMm}{11R}$$

### The Beauty of Mass Independence

Take a close look at our equation. The mass of the ball, $m$, appears on both sides. This means we can cancel it out completely!

This is a profound realization: the velocity required to reach a certain height is entirely independent of how heavy the object is. Whether it's a tennis ball or a massive spaceship, the required initial speed is exactly the same.

Let's multiply by 2 and take the square root to isolate $v_i$.

$$v_i = \sqrt{\frac{20GM}{11R}}$$

### The Escape Velocity Connection

The problem asks us to express this velocity in terms of the escape velocity, $v_e$. We know that the escape velocity from Earth's surface is given by a specific formula.

$$v_e = \sqrt{\frac{2GM}{R}}$$

We need to cleverly manipulate our expression for $v_i$ to reveal this hidden $v_e$. We can split the number 20 into $10 	imes 2$.

$$v_i = \sqrt{\frac{10}{11} 	imes \frac{2GM}{R}}$$

By separating the square roots, the magic happens.

$$v_i = \sqrt{\frac{10}{11}} 	imes \sqrt{\frac{2GM}{R}}$$

That second term is exactly our escape velocity!

$$v_i = \sqrt{\frac{10}{11}} v_e$$

### The Final Verdict

We are given that the velocity can be written in the form $v_i = \sqrt{\frac{x}{y}} v_e$. By comparing this with our derived expression, the mapping is crystal clear.

The numerator $x$ corresponds to 10, and the denominator $y$ corresponds to 11.

Therefore, the value of $x$ is **10**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Projection$

$Conservation of Mechanical Energy$

$Energy at the Surface$

$Energy at Maximum Height$

$Equating Initial and Final Energies$

$Solving for v_{i}$

$Expression for v_{i}$

$Introducing Escape Velocity$

$Relating v_{i} and v_{e}$

$Finding the value of x$

$The Way Forward$

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Setup

The Master Equation

Analyzing the Launch

Reaching the Peak

The Grand Equating

The Beauty of Mass Independence

The Escape Velocity Connection

The Final Verdict

Similar Questions

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:

The masses and radii of the Earth and the Moon are $M_{1}, R_{1}$ and $M_{2}, R_{2}$ respectively. Their centres are a distance $d$ apart. The minimum speed with which a particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is _________.

There is a crater of depth $\frac{R}{100}$ on the surface of the moon (radius $R$ ). A projectile is fired vertically upward from the crater with velocity, which is equal to the escape velocity $v$ from the surface of the moon. Find the maximum height attained by the projectile.

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $11 kms^{- 1}$ , the escape velocity from the surface of the planet would be

Two bodies, each of mass $M$ , are kept fixed with a separation $2 L$ . A particle of mass $m$ is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is $G$ . The correct statement(s) is (are)

If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth, is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Projection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy at the Surface

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy at Maximum Height

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Initial and Final Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for vi​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expression for vi​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Introducing Escape Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating vi​ and ve​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the value of x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward