A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it

Q: Why can't we just use the simple potential energy formula $\Delta U = mgh$ here?

Because the height $h$ is comparable to the radius of the planet $R$. The formula $\Delta U = mgh$ is an approximation valid only when $h \ll R$, where gravity can be assumed constant. Here, gravity drops to a quarter of its value, so we must use the exact potential energy formula.

Q: Does the angle of projection matter for the maximum height reached?

No! Since gravity is a conservative central force, the total mechanical energy depends only on the distance from the center of the planet. As long as the bullet does not collide with the planet's surface, the maximum height reached depends solely on the initial speed $v$, not the launch angle.

Q: How is the formula $\Delta U = \frac{mgh}{1 + h/R}$ derived?

It is derived from the fundamental gravitational potential energy formula $U = -\frac{GMm}{r}$. The potential energy at the surface is $U_A = -\frac{GMm}{R}$, and at height $h$ is $U_B = -\frac{GMm}{R+h}$. Subtracting these gives $\Delta U = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right) = \frac{GMm h}{R(R+h)} = \frac{mgh}{1 + h/R}$ since $g = \frac{GM}{R^2}$.

Q: What would change if we did not ignore atmospheric resistance?

If atmospheric resistance were present, some mechanical energy would be dissipated as heat. Consequently, the bullet would need a higher initial velocity $v$ to reach the same height $h = R$. This would increase $v$, making the ratio $v_{\text{esc}}/v$ smaller, which means $N$ would be less than 2.

Q: Why is the velocity at maximum height zero?

At the maximum height, the bullet momentarily stops before falling back down. Therefore, its instantaneous vertical velocity is zero, which means its kinetic energy at that point is also zero.

Question

A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is $\frac{1}{4}$th of its value at the surface of the planet. If the escape velocity from the planet is $v_{	ext{esc}} = v\sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere)

Sigma Percentile · Accepted Answer

### The Cosmic Cannonball: An Introduction

Imagine standing on the surface of a massive, airless planet.

You hold a powerful launcher, pointing it straight up into the starry abyss.

With a deafening roar, a bullet is fired vertically upwards with an initial velocity $v$.

As it climbs, it fights against the invisible, relentless pull of the planet's gravity.

It slows down, meter by meter, until it reaches its peak—a point of momentary stillness before the long fall back home.

This problem asks us to connect the dynamics of this vertical flight with one of the most fundamental escape parameters in astrophysics: the escape velocity.

Let's embark on this mathematical and physical journey to find the secret ratio $N$ that links these two velocities.

### Phase 1: Decoding the Gravity Gradient

In introductory physics, we often treat gravity as a constant force.

We write $F = mg$ and assume that the acceleration due to gravity is always $9.8 	ext{ m/s}^2$.

But when we deal with cosmic scales, where heights are comparable to the size of the planet itself, this approximation crumbles.

As we move away from the center of a spherical planet, the gravitational force weakens in accordance with Newton's inverse-square law.

The acceleration due to gravity $g'$ at a height $h$ above the surface of a planet of radius $R$ is given by the precise formula:

$$g' = \frac{GM}{(R+h)^2}$$

At the surface, where $h = 0$, the acceleration is simply:

$$g = \frac{GM}{R^2}$$

By dividing these two equations, we get the elegant relation showing how gravity scales with altitude:

$$g' = \frac{g}{\left(1 + \frac{h}{R}ight)^2}$$

This formula is our first major tool.

It tells us exactly how much the planet's grip loosens as the bullet climbs higher.

### Phase 2: The Geometry of the Peak

The problem gives us a crucial clue: at the maximum height $h$, the acceleration due to gravity is exactly $\frac{1}{4}$th of its value at the surface.

This means we can set $g' = \frac{g}{4}$.

Let's substitute this into our scaling formula:

$$\frac{g}{4} = \frac{g}{\left(1 + \frac{h}{R}ight)^2}$$

Canceling $g$ from both sides, we get:

$$\left(1 + \frac{h}{R}ight)^2 = 4$$

Taking the square root of both sides (and keeping the positive root since physical distances must be positive):

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

Subtracting $1$ from both sides yields:

$$\frac{h}{R} = 1 \implies h = R$$

What a beautiful and clean result!

The bullet rises to a height exactly equal to the radius of the planet.

This means at its peak, the bullet is at a distance of $2R$ from the center of the planet.

### Phase 3: The Symphony of Energy Conservation

Now that we know the geometry of the flight, how do we find the launch velocity $v$?

Since there is no atmosphere to dissipate energy through friction, the total mechanical energy of the bullet is conserved.

As the bullet climbs, its kinetic energy is converted entirely into gravitational potential energy.

Let's write down the conservation of energy between the launch point $A$ (on the surface) and the peak point $B$ (at height $h = R$):

$$K_A + U_A = K_B + U_B$$

At the peak, the bullet momentarily stops, so its final kinetic energy is zero:

$$K_B = 0$$

Therefore, the initial kinetic energy must equal the gain in potential energy:

$$K_A = U_B - U_A = \Delta U$$

The initial kinetic energy is:

$$K_A = \frac{1}{2}mv^2$$

The potential energy of a mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by:

$$U = -\frac{GMm}{r}$$

At the surface ($r = R$):

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

At the peak ($r = R + h = 2R$):

$$U_B = -\frac{GMm}{2R}$$

Let's calculate the gain in potential energy:

$$\Delta U = U_B - U_A = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}ight) = \frac{GMm}{2R}$$

Alternatively, we can use the standard formula for potential energy gain over large heights:

$$\Delta U = \frac{mgh}{1 + \frac{h}{R}}$$

Substituting $h = R$ into this formula:

$$\Delta U = \frac{mgR}{1 + 1} = \frac{mgR}{2}$$

Both methods yield the exact same result, showing the beautiful consistency of physics!

Now, let's equate the kinetic energy to this potential energy gain:

$$\frac{1}{2}mv^2 = \frac{mgR}{2}$$

Canceling the mass $m$ and the factor of $\frac{1}{2}$ from both sides, we get:

$$v^2 = gR \implies v = \sqrt{gR}$$

This is the precise launch velocity required to reach a height equal to the planet's radius.

### Phase 4: Connecting to Escape Velocity

Now, let's bring in the second player of our problem: the escape velocity $v_{	ext{esc}}$.

Escape velocity is the minimum speed required for an object to escape the gravitational pull of a planet completely and reach infinity with zero kinetic energy.

Using energy conservation from the surface to infinity ($r = \infty$):

$$K_{	ext{surface}} + U_{	ext{surface}} = K_{\infty} + U_{\infty}$$

Since the object just escapes, both its final kinetic energy and potential energy at infinity are zero:

$$\frac{1}{2}mv_{	ext{esc}}^2 - \frac{GMm}{R} = 0 + 0$$

Solving for $v_{	ext{esc}}$:

$$v_{	ext{esc}} = \sqrt{\frac{2GM}{R}}$$

Since $g = \frac{GM}{R^2}$, we can write $GM = gR^2$.

Substituting this in, we get the classic formula for escape velocity in terms of surface gravity:

$$v_{	ext{esc}} = \sqrt{2gR}$$

### Phase 5: The Final Synthesis

We are given the relation:

$$v_{	ext{esc}} = v\sqrt{N}$$

Let's substitute our expressions for both $v_{	ext{esc}}$ and $v$:

$$\sqrt{2gR} = \sqrt{gR} \cdot \sqrt{N}$$

Squaring both sides to remove the square roots:

$$2gR = N \cdot gR$$

Dividing both sides by $gR$ (since $gR 
eq 0$):

$$N = 2$$

And there it is!

The value of $N$ is exactly $2$.

This means the escape velocity is exactly $\sqrt{2}$ times the velocity required to reach a height equal to the planet's radius.

### Conclusion: The Beauty of Gravitational Mechanics

This problem is a classic example of the elegance of JEE physics.

It beautifully weaves together the concepts of gravitational field variation, work-energy theorem, and escape velocity.

By breaking down the problem step-by-step, we see that what initially looked like a complex orbital mechanics problem is actually a straightforward application of energy conservation.

Keep practicing, keep visualizing, and let the laws of physics guide your intuition!

Enter Numerical Value:

Visualized Solution

Visualizing the Launch from the Planet's Surface

Variation of Acceleration due to Gravity

Finding the Maximum Height $h$

Conservation of Mechanical Energy

Setting up the Energy Equation

Solving for Launch Velocity $v$

The Concept of Escape Velocity

Comparing Launch Velocity and Escape Velocity

Conclusion

Exploring Further: Launch Angle and Atmosphere

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Cannonball

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

Similar Questions

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 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:

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

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

Visualizing the Launch from the Planet's Surface

Variation of Acceleration due to Gravity

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

Conservation of Mechanical Energy

Setting up the Energy Equation

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

The Concept of Escape Velocity

Comparing Launch Velocity and Escape Velocity

Conclusion

Exploring Further: Launch Angle and Atmosphere

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Cannonball

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

Similar Questions

Finding the Maximum Height $h$

Solving for Launch Velocity $v$