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Chapter 10: Problem 82

The escape velocity from the earth is \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth would be (a) \(5.5 \mathrm{kms}^{-1}\) (b) \(11 \mathrm{kms}^{-1}\) (c) \(15.5 \mathrm{kms}^{-1}\) (d) \(22 \mathrm{kms}^{-1}\)

Short Answer

Expert verified
The escape velocity from the new planet is \(22 \mathrm{kms}^{-1}\) (Option d).

Step by step solution

01

Understand the Escape Velocity Formula

The escape velocity from a planet is given by the formula \( v = \sqrt{\frac{2 G M}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. Since the mean density \( \rho = \frac{M}{V} \), where \( V \) is the volume, and volume of a sphere \( V = \frac{4}{3} \pi R^3 \), the mass \( M = \rho \times \frac{4}{3} \pi R^3 \). This gives the escape velocity formula as \( v = \sqrt{\frac{2 G \rho \cdot \frac{4}{3} \pi R^3}{R}} \), or simplified as \( v = \sqrt{\frac{8 \pi G \rho R^2}{3}} \).
02

Set Expressions for Earth and the New Planet

For Earth, we have \( v_e = 11 \mathrm{kms}^{-1} = \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \). For the new planet (with twice the radius as Earth and the same density), the escape velocity \( v_p = \sqrt{\frac{8 \pi G \rho_e (2R_e)^2}{3}} \).
03

Simplify New Planet's Escape Velocity

Since \( R_p = 2R_e \) and \( \rho_p = \rho_e \), substituting in the new planet's escape velocity gives \( v_p = \sqrt{\frac{8 \pi G \rho_e (2R_e)^2}{3}} = \sqrt{\frac{8 \pi G \rho_e \cdot 4R_e^2}{3}} = 2 \times \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \).
04

Calculate and Compare the Results

Since \( \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} = v_e \), we have \( v_p = 2v_e = 2 \times 11 \mathrm{kms}^{-1} = 22 \mathrm{kms}^{-1} \). Thus, the escape velocity for the new planet is \( 22 \mathrm{kms}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Constant
Understanding the gravitational constant is crucial in the study of celestial mechanics. Denoted by the symbol \( G \), the gravitational constant is a key value that helps measure the force of attraction between two bodies due to gravity.
It plays an essential role in determining how objects move in a gravitational field. - **Value of \( G \):** Typically, \( G \) has a value of approximately \( 6.674 \, \times \, 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). This very small value highlights how weak the gravitational force is compared to other fundamental forces. - **Importance in Formulas:** In the formula for escape velocity \( v = \sqrt{\frac{2GM}{R}} \), \( G \) allows us to calculate the speed needed for an object to break free from a planet鈥檚 gravitational pull. - **Effect on Planetary Motion:** The gravitational constant is crucial for calculating the potential energy and movement of objects in orbits around larger bodies, like planets around stars or moons around planets.By understanding \( G \), students can better grasp how planets, moons, and other celestial bodies interact through the fundamental force of gravity.
Planetary Density
Planetary density is a measure of how much mass is contained within a planet鈥檚 volume.
It is intimately connected with the escape velocity and affects how strongly a planet's gravity pulls on nearby objects. - **Definition:** The density \( \rho \) of a planet is given by the formula \( \rho = \frac{M}{V} \), where \( M \) is the mass and \( V \) is the volume. - **Relation with Volume:** Since planets are mostly spherical, their volume \( V \) is calculated using the formula \( \frac{4}{3} \pi R^3 \). Therefore, density is indirectly connected to the radius through volume. - **Impact on Escape Velocity:** In the formula \( v = \sqrt{\frac{8 \pi G \rho R^2}{3}} \), density directly influences the required escape velocity. The denser a planet, the higher the escape velocity required because mass is a component of gravitational attraction.By understanding density, one can predict how easily or difficult it is for objects to reach escape velocity and leave a planet's gravitational influence.
Radius of a Planet
The radius of a planet is a crucial factor that influences numerous other characteristics, such as gravitational pull and escape velocity.
It can drastically alter the conditions on a planet鈥檚 surface. - **Definition and Measurement:** A planet's radius is the distance from its center to its surface. This measurement is essential in various calculations involving gravitational force. - **Connection with Escape Velocity:** In the escape velocity formula \( v = \sqrt{\frac{2GM}{R}} \), the radius \( R \) plays a fundamental role. A larger radius requires a higher escape velocity assuming density remains constant. - **Influence on Surface Gravity:** Gravity at a planet's surface is also determined by its radius and mass. A larger planet with the same mass as a smaller one will have weaker surface gravity.Understanding the radius helps in predicting physical and potentially even life-supporting conditions on different planets, making it a fundamental parameter in the field of astronomy.

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Most popular questions from this chapter

A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clock run at the same rate on earth. On a planet having the same density as earth but twice the radius, (a) \(S\) will run faster than \(P\) (b) \(P\) will run faster than \(S\) (c) both will run at the same rate as on the earth (d) both will run at the same rate which will be different from that on the earth

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is (a) \(32000 \mathrm{~km}\) (b) \(36000 \mathrm{~km}\) (c) \(6400 \mathrm{~km}\) (d) \(4800 \mathrm{~km}\)

At a given place where, acceleration due to gravity is \(g \mathrm{~ms}^{-2}\), a sphere of lead of density \(d \mathrm{kgm}^{-3}\) is gently released in a column of liquid of density \(\rho \mathrm{kgm}^{-3}\). If \(d>\rho\), the sphere will (a) fall vertically with an acceleration of \(\mathrm{g} \mathrm{ms}^{-2}\) (b) fall vertically with no acceleration (c) fall vertically with an acceleration \(g\left(\frac{d-\rho}{d}\right)\) (d) fall vertically with an acceleration \(\rho / d\)

A satellite orbits the earth at a height of \(400 \mathrm{~km}\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 \mathrm{~kg}\), mass of the earth \(=6.0 \times 10^{24} \mathrm{~kg}\), radius of the earth \(=6.4 \times 10^{6} \mathrm{~m}, G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\). (a) \(5.2 \times 10^{10} \mathrm{~J}\) (b) \(3 \times 10^{6} \mathrm{~J}\) (c) \(4 \times 10^{6} \mathrm{~J}\) (d) \(6 \times 10^{9} \mathrm{~J}\)

Two satellites \(S_{1}\) and \(S_{2}\) revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are \(1 \mathrm{~h}\) and \(8 \mathrm{~h}\) respectively. The radius of orbit of \(S_{1}\) is \(10^{4} \mathrm{~km}\). When \(S_{2}\) is closest to \(S_{1}\), the speed of \(S_{2}\) relative to \(S_{1}\) is (a) \(\pi \times 10^{4} \mathrm{kmh}^{-1}\) (b) \(2 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (c) \(3 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (d) \(4 \pi \times 10^{4} \mathrm{kmh}^{-1}\)
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