Gravitational acceleration on the surface of a planet is (sqrt(6)/11)g, where g is the gravitational

Q: Why did we assume the planet is a uniform sphere?

In standard JEE physics problems, unless specified otherwise, planets are assumed to be uniform spheres. This allows us to use the standard volume formula $V = \frac{4}{3}\pi R^3$ and a constant density $\rho$ to relate mass and radius.

Q: Can we use $v_e = \sqrt{\frac{2GM}{R}}$ directly without converting to $g$?

Yes, but since we are given the ratio of surface gravities ($g_p/g_e$) rather than masses, converting the formula to use $g$ and $\rho$ is much faster and less prone to algebraic errors.

Q: Why is the density ratio inverted under the square root in the final ratio?

Because our scaling law is $v_e \propto \frac{g}{\sqrt{\rho}}$. Since density $\rho$ is in the denominator, the ratio of escape velocities $v_{e,p}/v_{e,e}$ must feature the inverse density ratio $\sqrt{\rho_e/\rho_p}$.

Q: What if the density of the planet was given as a mass ratio instead?

If mass and density ratios were both given, we could find the radius ratio directly from $M = \frac{4}{3}\pi R^3 \rho \implies R \propto \left(\frac{M}{\rho}\right)^{1/3}$, and then compute escape velocity using $v_e \propto \sqrt{\frac{M}{R}}$.

Q: Is escape velocity dependent on the mass of the projected object?

No, escape velocity depends only on the mass and radius of the planet (or its surface gravity and radius). The mass of the projected object cancels out when equating kinetic energy to gravitational potential energy.

Question

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} g$, where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11	ext{ km s}^{-1}$, the escape speed on the surface of the planet in $	ext{ km s}^{-1}$ will be

Sigma Percentile · Accepted Answer

### The Cosmic Setup

Imagine standing on the surface of a newly discovered planet.

You want to launch a probe into deep space so that it never returns.

To do this, you need to know the **escape velocity** of this planet.

We are given that the planet's surface gravity is a fraction of Earth's:

$$g_p = \frac{\sqrt{6}}{11} g_e$$

And its average mass density is two-thirds of Earth's:

$$ho_p = \frac{2}{3} ho_e$$

Knowing that Earth's escape velocity is $11	ext{ km/s}$, how do we find the escape velocity of this planet without knowing its mass or radius directly?

Let's embark on a beautiful journey of scaling laws to solve this elegantly.

### Unveiling the Proportionality

Let's start with the fundamental definition of acceleration due to gravity on the surface of a spherical body:

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

Since we are given the density $ho$ rather than the mass $M$, we must express mass in terms of density and radius.

Assuming a uniform spherical distribution, the mass is:

$$M = ho \cdot V = ho \left(\frac{4}{3}\pi R^3ight)$$

Substituting this back into our gravity equation yields:

$$g = \frac{G}{R^2} \left(\frac{4}{3}\pi R^3 hoight) = \frac{4}{3}\pi G ho R$$

This reveals a highly intuitive physical relationship: **surface gravity is directly proportional to the product of density and radius**.

$$g \propto ho R$$

Since we do not know the radius of the planet, we can rearrange this proportionality to express the radius in terms of gravity and density:

$$R \propto \frac{g}{ho}$$

### The Power of Scaling Laws

Now, let's look at the escape velocity formula.

The minimum speed required to escape a gravitational pull is:

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

Since $g = \frac{GM}{R^2}$, we can rewrite this as:

$$v_e = \sqrt{2gR}$$

This means the escape velocity scales as:

$$v_e \propto \sqrt{gR}$$

Now, let's substitute our expression for radius $R \propto \frac{g}{ho}$ into this scaling relation:

$$v_e \propto \sqrt{g \left(\frac{g}{ho}ight)} \implies v_e \propto \frac{g}{\sqrt{ho}}$$

This is our master scaling law!

It tells us that escape velocity is directly proportional to surface gravity and inversely proportional to the square root of density.

### The Elegant Cancellation

Let's set up the ratio of the escape velocity of the planet to that of Earth:

$$\frac{v_{e,p}}{v_{e,e}} = \left(\frac{g_p}{g_e}ight) \sqrt{\frac{ho_e}{ho_p}}$$

Now, substitute the given ratios:

$$\frac{g_p}{g_e} = \frac{\sqrt{6}}{11}$$

$$\frac{ho_p}{ho_e} = \frac{2}{3} \implies \frac{ho_e}{ho_p} = \frac{3}{2}$$

Plugging these in:

$$\frac{v_{e,p}}{v_{e,e}} = \left(\frac{\sqrt{6}}{11}ight) \sqrt{\frac{3}{2}}$$

We can combine the square roots:

$$\frac{v_{e,p}}{v_{e,e}} = \frac{1}{11} \sqrt{6 	imes \frac{3}{2}} = \frac{1}{11} \sqrt{9} = \frac{3}{11}$$

What an incredibly clean result!

The ratio of escape velocities is exactly $\frac{3}{11}$.

### Final Calculation

We are given that Earth's escape velocity is $v_{e,e} = 11	ext{ km/s}$.

Therefore, the escape velocity of the planet is:

$$v_{e,p} = v_{e,e} 	imes \frac{3}{11} = 11 	imes \frac{3}{11} = \mathbf{3	ext{ km/s}}$$

The final answer is **3**.

Enter Numerical Value:

Visualized Solution

Visualizing Earth vs. Planet

Relating Gravity, Density, and Radius

Expressing Mass as Volume times Density

Finding the Radius Ratio

Formula for Escape Velocity

Combining Proportionalities

Setting up the Ratio

Substituting the Given Ratios

Simplifying the Radical Expression

Calculating the Final Escape Velocity

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Setup

Unveiling the Proportionality

The Power of Scaling Laws

The Elegant Cancellation

Final Calculation

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