A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 time

Q: Why doesn't the escape velocity depend on the mass of the rocket?

Escape velocity is derived by equating Kinetic Energy ($\frac{1}{2}mv^2$) to Gravitational Potential Energy ($\frac{GMm}{R}$). The mass of the object ($m$) appears on both sides of the equation and cancels out. Therefore, whether you are launching a pebble or a massive starship, the required escape velocity is exactly the same!

Q: What if the planet was 10 times more massive but had the SAME density as Earth?

If the density was the same, the volume would have to be 10 times larger. Since Volume $\propto R^3$, the radius would be $10^{1/3}$ times larger. The escape velocity would then scale as $\sqrt{M/R} = \sqrt{10 / 10^{1/3}} = 10^{1/3} \approx 2.15$ times Earth's escape velocity, not 10 times.

Q: Does the angle at which we launch the rocket matter for escape velocity?

No! Escape velocity is a scalar quantity derived from energy conservation, not a vector calculation. As long as the rocket is not pointed directly into the ground and doesn't hit any obstacles, an initial speed of $v_e$ in any outward direction will allow it to escape to infinity.

Q: Where does the $11 \text{ km/s}$ value for Earth come from?

It comes from plugging Earth's mass ($M \approx 5.97 \times 10^{24} \text{ kg}$) and radius ($R \approx 6.37 \times 10^6 \text{ m}$) into the formula $v = \sqrt{\frac{2GM}{R}}$. Alternatively, using $g = \frac{GM}{R^2}$, you can write $v = \sqrt{2gR}$, which gives the same result.

Q: Can escape velocity ever be greater than the speed of light?

Yes! If a mass is compressed into a small enough radius (known as the Schwarzschild radius), the escape velocity equals or exceeds the speed of light ($c$). Nothing, not even light, can escape. This is the exact definition of a Black Hole.

Question

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 $11 	ext{ kms}^{-1}$, the escape velocity from the surface of the planet would be

Options:

(A) $1.1 	ext{ kms}^{-1}$

(B) $11 	ext{ kms}^{-1}$

(C) $110 	ext{ kms}^{-1}$

(D) $0.11 	ext{ kms}^{-1}$

Sigma Percentile · Accepted Answer

### The Cosmic Prison: Escaping a Super-Dense Planet

Imagine you are an astronaut tasked with launching a rocket from a newly discovered exoplanet. The mission briefing gives you two terrifying pieces of information: this planet is **10 times more massive** than Earth, and its radius is **10 times smaller**.

Before you even start your engines, you need to know one critical number: the escape velocity. How fast do you need to go to break free from this planet's gravitational grip forever? Let's break down the physics behind this cosmic challenge.

### The Master Equation: Escape Velocity

To escape a planet's gravitational field, an object must be launched with enough kinetic energy to overcome the planet's gravitational potential energy. By equating the kinetic energy at the surface to the potential energy well, we derive the classic formula for escape velocity:

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

Here, $G$ is the universal gravitational constant, $M$ is the mass of the planet, and $R$ is its radius. Notice that the escape velocity depends *directly* on the square root of the mass and *inversely* on the square root of the radius.

For our home planet, Earth, we know the escape velocity is approximately $11 	ext{ km/s}$. Let's call this $v_e$.

### Scaling the Universe

Now, let's look at our distant, super-dense planet. We are given its parameters relative to Earth:

$$M_p = 10 M_e$$

$$R_p = \frac{R_e}{10}$$

We can substitute these scaled values directly into our escape velocity formula to find the planet's escape velocity, $v_p$:

$$v_p = \sqrt{\frac{2G(10M_e)}{\left(\frac{R_e}{10}ight)}}$$

### The Final Calculation

This is where the algebra reveals the planet's true nature. The fraction in the denominator flips, sending that $10$ up to multiply with the $10$ in the numerator:

$$v_p = \sqrt{\frac{100 	imes 2GM_e}{R_e}}$$

We can pull the $100$ out of the square root as a $10$:

$$v_p = 10 \sqrt{\frac{2GM_e}{R_e}}$$

Notice that the term left inside the square root is exactly the formula for Earth's escape velocity, $v_e$! Therefore, we can elegantly write:

$$v_p = 10 v_e$$

Since we know $v_e = 11 	ext{ km/s}$, the final calculation is trivial:

$$v_p = 10 	imes 11 	ext{ km/s} = 110 	ext{ km/s}$$

To escape this planet, your rocket would need to reach a staggering **$110 	ext{ km/s}$**. This makes perfect physical sense: packing 10 times the mass into a tiny radius creates an incredibly deep gravitational well. In fact, the density of this planet would be $10,000$ times that of Earth!

Animated Solution for Physics - Gravitation: 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

Select Answer:

Visualized Solution

$M_{p} = 10 M_{e}, R_{p} = \frac{R _{e}}{10}$

$v = \frac{2 GM}{R}$

$v_{p} = \frac{2 G M _{p}}{R _{p}}$

$v_{p} = 10 \frac{2 G M _{e}}{R _{e}}$

$v_{p} = 110 km/s$

$Density ρ \propto \frac{M}{R ^{3}}$

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Prison

The Master Equation

Scaling the Universe

The Final Calculation

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