Two spherical planets P and Q have the same uniform density rho, masses M_P and M_Q, and surface are

Q: Why does escape velocity increase with radius if the formula $v = \sqrt{2GM/R}$ has radius in the denominator?

While it is true that $R$ is in the denominator, the mass $M$ in the numerator is not constant; it depends on the volume of the planet. Since mass is proportional to $R^3$ for a constant density, substituting $M \propto R^3$ into the formula yields $v \propto \sqrt{R^3/R} = R$. Thus, for planets of the same density, a larger radius means a much larger mass, which ultimately increases the escape velocity.

Q: Can we solve this problem without assuming the planets are spherical?

No, the formulas for surface area ($4\pi R^2$), volume ($\frac{4}{3}\pi R^3$), and gravitational potential energy on the surface ($-\frac{GMm}{R}$) strictly assume spherically symmetric mass distributions.

Q: How did we get $M_Q = 8M_P$ from the radius ratio?

Since both planets have the same uniform density $\rho$, their masses are directly proportional to their volumes. The volume of a sphere scales with the cube of its radius ($V \propto R^3$). Since $R_Q = 2R_P$, we have $M_Q \propto (2R_P)^3 = 8R_P^3 \propto 8M_P$.

Q: Is the escape velocity independent of the mass of the projected object?

Yes, the escape velocity formula $v = \sqrt{2GM/R}$ does not contain the mass of the projected object $m$. It only depends on the mass and radius of the planet.

Q: What if the densities of the planets were different?

If the densities were different, we could not use the simple proportionality $v \propto R$. We would have to calculate the escape velocity using the individual masses and radii directly using $v \propto \sqrt{M/R}$.

Question

Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_P$ and $M_Q$, and surface areas $A$ and $4A$, respectively. A spherical planet $R$ also has uniform density $\rho$ and its mass is $(M_P + M_Q)$. The escape velocities from the planets $P, Q$, and $R$, are $v_P, v_Q$, and $v_R$, respectively. Then

Options:

(A) $v_Q > v_R > v_P$

(B) $v_R > v_Q > v_P$

(C) $v_R/v_P = 3$

(D) $v_P/v_Q = 1/2$

Sigma Percentile · Accepted Answer

### Introduction: The Cosmic Escape Imagine standing on the surface of a distant planet, holding a projectile. If you throw it upwards, gravity pulls it back. But what if you throw it with such immense speed that it breaks free from the planet's gravitational grip forever? This critical threshold speed is known as the **escape velocity**. In this problem, we are introduced to three spherical planets: $P$, $Q$, and $R$. All three share a common trait—they have the exact same **uniform density** $ ho$. However, they differ in their sizes, masses, and surface areas. Our mission is to compare their escape velocities and find the correct relationships among them. ### The Physics of Breaking Free To find the escape velocity, we must look at the conservation of mechanical energy. For a projectile of mass $m$ to just escape to infinity from the surface of a planet of mass $M$ and radius $R$, its total mechanical energy at infinity must be at least zero. By equating the initial kinetic and potential energy on the surface to the energy at infinity, we get: $$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0$$ Solving for $v$, we arrive at the classic escape velocity formula: $$v = \sqrt{\frac{2GM}{R}}$$ Here, $G$ is the universal gravitational constant. This formula tells us that escape velocity depends on both the mass and the radius of the planet. ### The Density Connection: A Clever Simplification Since we are given that all three planets have the same uniform density $ ho$, we can express the mass $M$ of a planet in terms of its radius $R$ and density $ ho$: $$M = ext{Volume} imes ext{Density} = \frac{4}{3}\pi R^3 ho$$ Now, let us substitute this expression for mass back into our escape velocity formula: $$v = \sqrt{\frac{2G}{R} \left(\frac{4}{3}\pi R^3 ho ight)}$$ Simplifying the expression inside the square root, we get: $$v = R\sqrt{\frac{8}{3}\pi G ho}$$ This is a beautiful and powerful result! Since $G$ and $ ho$ are constant for all three planets, the entire term inside the square root is a constant. Therefore, for planets of uniform density, the escape velocity is **directly proportional to the radius**: $$v \propto R$$ This single proportionality reduces our complex gravitational problem into a straightforward geometric comparison of the radii of the three planets. ### Analyzing Planet P and Planet Q We are given that the surface areas of planets $P$ and $Q$ are $A$ and $4A$, respectively. Since the surface area of a sphere is given by $4\pi R^2$, we can write: $$\frac{A_Q}{A_P} = \frac{4\pi R_Q^2}{4\pi R_P^2} = \frac{4A}{A} = 4$$ Taking the square root on both sides, we find the ratio of their radii: $$\frac{R_Q}{R_P} = 2 \implies R_Q = 2R_P$$ Let us define the radius of planet $P$ as $r$. Then, the radius of planet $Q$ is simply $2r$. Using our proportionality $v \propto R$, the ratio of their escape velocities is: $$\frac{v_P}{v_Q} = \frac{R_P}{R_Q} = \frac{r}{2r} = \frac{1}{2}$$ This immediately confirms that **option (d)** is correct! Now, let us also find the relationship between their masses. Since mass scales with the cube of the radius ($M \propto R^3$ for constant density), we have: $$\frac{M_Q}{M_P} = \left(\frac{R_Q}{R_P} ight)^3 = 2^3 = 8$$ If we let the mass of planet $P$ be $M$, then the mass of planet $Q$ is $8M$. ### Introducing Planet R: The Sum of Worlds Next, we are introduced to planet $R$, which also has the same uniform density $ ho$. Its mass $M_R$ is given as the sum of the masses of $P$ and $Q$: $$M_R = M_P + M_Q = M + 8M = 9M$$ Since planet $R$ has the same density, we can find its radius $R_R$ by comparing it to planet $P$: $$\frac{M_R}{M_P} = \left(\frac{R_R}{R_P} ight)^3 = \frac{9M}{M} = 9$$ Taking the cube root of both sides: $$R_R = 9^{1/3} R_P = 9^{1/3} r$$ Using a quick approximation, since $2^3 = 8$, the cube root of $9$ is slightly greater than $2$: $$9^{1/3} \approx 2.08$$ Thus, the radius of planet $R$ is approximately $2.08r$. ### The Final Comparison: Ordering the Velocities Now we have the radii of all three planets expressed in terms of $r$: * Radius of $P$: $R_P = r$ * Radius of $Q$: $R_Q = 2r$ * Radius of $R$: $R_R = 9^{1/3}r \approx 2.08r$ Comparing these values, we get the clear ordering: $$r < 2r < 2.08r \implies R_P < R_Q < R_R$$ Since escape velocity is directly proportional to the radius ($v \propto R$), their escape velocities must follow the exact same order: $$v_P < v_Q < v_R \implies v_R > v_Q > v_P$$ This perfectly matches **option (b)**! ### Conclusion: The Power of Proportionality By translating the physical laws of gravitation into simple proportionalities, we avoided tedious calculations and solved the problem with elegant logic. We found that: 1. The ratio of escape velocities of $P$ and $Q$ is $v_P/v_Q = 1/2$. 2. The correct ordering of escape velocities is $v_R > v_Q > v_P$. Thus, the correct options are **(b)** and **(d)**.

Select Answer:

Visualized Solution

Visualizing the Three Planets $P$ , $Q$ , and $R$

The Physics of Escape Velocity

Relating Escape Velocity to Density

Finding the Radius of Planet $Q$

Calculating the Mass of Planet $Q$

Comparing Escape Velocities of $P$ and $Q$

Analyzing Planet $R$

Finding the Radius of Planet $R$

Comparing Escape Velocity of $R$ with $P$ and $Q$

Establishing the Final Ordering

Summary of Correct Options

The Sigma Insight: Gravitational Potential and Potential Energy

Introduction

The Physics of Breaking Free

The Density Connection

Analyzing Planet P and Planet Q

Introducing Planet R

The Final Comparison

Conclusion

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