Problem 707 - JEE Physics

Question

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is \(\mathrm{R}\), the radius of planet would be (A) \(2 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(1 / 4 \mathrm{R}\) (D) \(\mathrm{R} / 2\)

Step-by-Step Solution

Answer

The planet's radius is half the Earth's radius, which is represented as (D) \(\mathrm{R} / 2\).

Step 1: Write down the gravitational force formula

The formula for gravitational force between two objects is \( F = G * \frac{m_1 * m_2}{r^2}\), where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers.

Step 2: Write down the density formula

The formula for density is \(ρ = \frac{m}{V}\), where ρ is the density, m is the mass, and V is the volume. Since both Earth and the planet are assumed to be spheres, their volumes can be represented by the formula for the volume of a sphere, which is \(V = \frac{4}{3}πr^3\).

Step 3: Set up equations for Earth and the planet

Let's denote Earth's density as \(ρ_e\) and the planet's density as \(ρ_p\). We know that \(ρ_p = 2ρ_e\). We also know that the gravitational forces on their surfaces are the same. Therefore, we can set up two equations: 1. For Earth: \[F_e = G * \frac{m_e * m}{r_e^2}\] 2. For the planet: \[F_p = G * \frac{m_p * m}{r_p^2}\] Since \(F_e = F_p\), let's set these two equations equal to each other and eliminate G and m: \(G * \frac{m_e * m}{r_e^2} = G * \frac{m_p * m}{r_p^2}\)

Step 4: Express the masses of Earth and the planet in terms of their densities

From the density formulas, we have: 1. \(m_e = ρ_e * V_e = ρ_e * \frac{4}{3}πr_e^3\) 2. \(m_p = ρ_p * V_p = (2ρ_e) * \frac{4}{3}πr_p^3\) Now, substitute these expressions for the masses into the equation from Step 3: \(\frac{\rho_e * \frac{4}{3}πr_e^3 * m}{r_e^2} = \frac{(2\rho_e) * \frac{4}{3}πr_p^3 * m}{r_p^2}\)

Step 5: Solve for the planet's radius in terms of Earth's radius

First, cancel out the common terms from the equation: \(\frac{r_e^3 * \rho_e}{r_e^2} = \frac{2 * r_p^3 * \rho_e}{r_p^2}\) Next, cancel out the remaining common terms: \(r_e = 2 * r_p\) Now, solve for the planet's radius, \(r_p\): \(r_p = \frac{r_e}{2}\) So the correct answer is (D) \(\mathrm{R} / 2\). The planet's radius is half the Earth's radius.

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept that describes the force of attraction between two masses. This force is given by Newton's law of universal gravitation, represented by the formula: \ \[ F = G \frac{m_1 m_2}{r^2} \] where:

• \( F \) is the gravitational force, measured in Newtons,
• \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \),
• \( m_1 \) and \( m_2 \) are the masses of the two objects,
• \( r \) is the distance between the centers of the two masses.

This formula highlights that gravitational force increases with greater mass and decreases with increased distance between objects. For planets, this force is what gives us weight on their surfaces, affecting phenomena like tides and satellite orbits.

Density Formula

Density is a key property that relates mass to volume in a substance or object. It is defined by the formula: \ \[ \rho = \frac{m}{V} \] where:

• \( \rho \) is the density,
• \( m \) is the mass,
• \( V \) is the volume.

In our planet density scenario, the given density of a planet is twice that of Earth. Therefore, if we know the Earth's density, the planet’s density can be directly calculated as \( 2\rho_e \).
This concept is crucial because it gives insight into the internal structure and composition of a celestial body, helping us understand surface conditions like gravity.

Sphere Volume Calculation

Understanding the volume of a sphere is essential for various applications, including planetary sciences. The volume of a sphere is calculated with the following formula: \ \[ V = \frac{4}{3} \pi r^3 \] where:

• \( V \) is the volume,
• \( r \) is the radius of the sphere.

For planets, which are approximately spherical, knowing the radius allows scientists to calculate their volume. This volume plays a vital role when considering density and comparing physical properties between different planets.
When solving problems involving planetary volume, it becomes possible to relate volume changes to other factors like mass and density, which are crucial in gravitational physics.

Equal Surface Gravity Conditions

The concept of equal surface gravity conditions refers to having the same gravitational pull on the surface of two different celestial bodies. Here, both Earth and an unknown planet have such a condition despite having differing densities and radii.
When replacement is performed in our calculations, it involves equating the gravitational force experienced by objects on their surfaces. Normally for planets, this is equated as: \ \[ \frac{G m_e m}{r_e^2} = \frac{G m_p m}{r_p^2} \] where the subscript \( e \) denotes Earth and \( p \) denotes the planet.
By maintaining equal gravitational forces while acknowledging differences in density, you can solve for unknown planetary characteristics, such as radius, as in this example.
The example demonstrates how differing size and density can yet result in equal gravitational experiences at the surface. This understanding aids significantly in the fields of astronomy and geophysics.