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Chapter 6: Problem 707

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) \(\mathrm{R} / 2\)

Short Answer

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

Step by step solution

01

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.
02

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 \(V = \frac{4}{3}πr^3\).
03

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: \(G * \frac{m_e * m}{r_e^2} = G * \frac{m_p * m}{r_p^2}\)
04

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: \(\frac{\rho_e * \frac{4}{3}πr_e^3 * m}{r_e^2} = \frac{(2\rho_e) * \frac{4}{3}πr_p^3 * m}{r_p^2}\)
05

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.

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Most popular questions from this chapter

A planet moving along an elliptical orbit is closest to the sun at a distance \(\mathrm{r}_{1}\) and farthest away at a distance of \(\mathrm{r}_{2}\). If \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) are the liner velocities at these points respectively, then the ratio \(\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) is \(\ldots \ldots \ldots \ldots\) (A) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (C) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (D) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)^{2}\)

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

4 A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is $\ldots \ldots \ldots \ldots$ hours. (A) \(4.8\) (B) \(48 \sqrt{2}\) (C) 24 (D) \(24 \sqrt{2}\)

He period of revolution of planet \(\mathrm{A}\) around the sun is 8 times that of \(\mathrm{B}\). The distance of A from the sun is how many times greater than that of \(\mathrm{B}\) from the sun. (A) 2 (B) 3 (C) 4 (D) 5

The acceleration due to gravity near the surface of a planet of radius \(\mathrm{R}\) and density \(\mathrm{d}\) is proportional to (A) \(\mathrm{d} / \mathrm{R}^{2}\) (B) \(\mathrm{d} \mathrm{R}^{2}\) (C) \(\mathrm{dR}\) (D) \(\mathrm{d} / \mathrm{R}\)
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