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Chapter 13: Problem 31

The star Rho \({ }^{1}\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho \({ }^{1}\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho " Cancri?

Short Answer

Expert verified
(a) The orbital speed of the planet is approximately \( 31356.5 \) m/s. (b) The orbital period of the planet is approximately 0.52 Earth years.

Step by step solution

01

Calculate Orbital Speed

First, let's calculate the orbital speed for the planet revolving around Rho \( ^{1} \) Cancri. Using the formula for orbital speed \( v = \sqrt{\frac{GM}{r}} \), where G is the gravitational constant, M is the mass of the star, and r is the radius of the orbit. Given that the mass of the star is equal to 0.85 solar masses, we convert it into kilograms using the known mass of the sun \( M_{sun} = 1.989*\( 10^30 \) kg \). Similarly, as the radius of planet's orbit is 0.11 times Earth's orbit (also known as 1 Astronomical Unit - AU), we can convert it into meters using 1 AU = \( 1.496*10^{11} \) m.
02

Compute Orbital Speed

After computations, the orbital speed \( v = \sqrt{\frac{(6.67*10^-11 m^{3}kg^{-1}s^{-2})* (0.85*1.989*10^{30} kg)}{(0.11*(1.496*10^{11})m)}} = 31356.5 \) m/s.
03

Determine Orbital Period

Now, to calculate the planet's orbital period around Rho \( ^{1} \) Cancri, we will use Kepler's Third Law, formulated as \( T^2 = \left( \frac{4\pi^2}{GM} \right) * r^3 \), where T is the period, r is the radius of the orbit, G is the Gravitational Constant and M is the mass of the star.
04

Compute Orbital Period

Upon applying the given data, the period T is calculated as \( T = \sqrt{\left( \frac{4\pi^2}{6.67*10^-11 m^{3}kg^{-1}s^{-2}} \right) * (0.85*1.989*10^{30} kg)*(0.11*1.496*10^{11}m)^3} = 32804896.5 \) seconds or approximately 0.52 Earth Years.

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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, denoted as G, is fundamental in orbital mechanics. This constant is crucial in calculating the forces between two objects due to gravity. It's a key part of Newton's law of universal gravitation and appears in the equations we use to determine the movements of celestial bodies.

The value of the gravitational constant is approximately \( 6.67\times10^{-11} \text{m}^{3}\text{kg}^{-1}\text{s}^{-2} \), and it tells us the strength of gravity for given masses over a certain distance. For example, when calculating the orbital speed of a planet around a star, we use G to understand the influence of the star's mass on the planet's motion.
Kepler's Third Law
Kepler's Third Law is a glorious testament to the harmonious patterns found in the cosmos. It states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit, assuming the mass of the planet is much less than the mass of the star.

In mathematical terms, this is expressed as \({T^2 \propto r^3}\) or, with the inclusion of the gravitational constant and the mass of the central body, \(T^2 = \left( \frac{4\pi^2}{GM} \right) \times r^3\). This principle allows us to predict how long it will take for a celestial body to complete one full orbit around its star, which is quite impressive given the vast scale of space we're dealing with.
Orbital Period
The orbital period is the time a celestial object takes to complete one orbit around another object. It's an essential concept in understanding the motion of planets, moons, and artificial satellites. The period depends on the distance from the object it orbits and its speed.

Using our exercise as an example, when the mass of a star and the radius of a planet's orbit are known, we apply Kepler's Third Law to find the time the planet takes to travel around the star. This is very useful for astronomers who look for patterns in the movement of planets, which can reveal important information about their characteristics and the structure of their solar systems.
Orbital Speed
Orbital speed refers to the speed at which a planet or moon travels as it orbits a larger body, like a star or planet. It's determined by the balance of gravitational pull exerted by the larger body and the inertia of the object in motion.

When we calculate the orbital speed (like we did in the step-by-step solution), we're essentially finding out how fast the planet moves to maintain its circular orbit around the star Rho \( ^{1} \) Cancri. This speed is crucial because it balances the gravitational force; any slower, and the planet would spiral toward the star, any faster, and it would escape into space. Fascinating, isn't it?

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Most popular questions from this chapter

A thin, uniform rod has length \(L\) and mass \(M\). A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.38). (a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\). (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B.) (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L\)

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think we should land a vehicle there to search for life. Before launching it, we would want to test such a lander under the gravity conditions at the surface of Europa. One way to do this is to put the lander at the end of a rotating arm in an orbiting earth satellite. If the arm is \(4.25 \mathrm{~m}\) long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa? The mass of Europa is \(4.80 \times 10^{22} \mathrm{~kg}\) and its diameter is \(3120 \mathrm{~km}\)

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Two spherically symmetric planets with no atmosphere have the same average density, but planet \(B\) has twice the radius of planet \(A\). A small satellite of mass \(m_{A}\) has period \(T_{A}\) when it orbits planet \(A\) in a circular orbit that is just above the surface of the planet. A small satellite of mass \(m_{B}\) has period \(T_{B}\) when it orbits planet \(B\) in a circular orbit that is just above the surface of the planet. How does \(T_{B}\) compare to \(T_{A} ?\) with a density of \(2500 \mathrm{~kg} / \mathrm{m}^{3} ?\)

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