Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2MEarth) orbits at a distance of 1 AU from the star. What is the orbital period of this planet? Hint: Think about how the mass of the Sun compares with the mass of the Earth. a. 3 months b. 6 months
c. 1 year d. 2 years
e. It would not be able to orbit at this distance.

Answers

Answer 1

The correct answer is option D.2 years

What is Kepler's third law of planetary motion?

According to Kepler's Third Law of Planetary Motion, T² is proportional to r³, where T is the period of revolution of the planet and r is the distance between the planet and the star.

In order to solve for T,

AU = 1

Astronomical Unit = the average distance between the Earth and the Sun = 149.6 million kilometres

Therefore, the planet is orbiting at a distance of 149.6 million kilometres from the star.

Substituting the values of r and solving for

T².T² ∝ r³T² ∝ (149.6)³T²

= (149.6)³T²

= 3.522 x 10¹²T

= √3.522 x 10^¹²T

= 1.87 x 10⁶ seconds

T = 31,100 minutes

T = 518 hours

T = 21.6 days

T = 2 years

Therefore, the orbital period of the planet with twice the mass of Earth orbiting at a distance of 1 AU from a star with the same mass as the Sun is 2 years.

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Related Questions

charge q1 is distance s from the negative plate of a parallel-plate capacitor. charge is distance 2s from the negative plate. what is the ratio of their potential energies?

Answers

The electric potential energy, U, of two point charges is given by the equation, U = kq1q2/r where k is Coulomb's constant, q1 and q2 are the charges and r is the distance between the two charges. Now, let's solve the question using this equation. There are two charges, q1 and q2, and a parallel plate capacitor between them. The distance of q1 from the negative plate is s, and the distance of q2 from the negative plate is 2s. The charges have the same magnitude of charge, so let's assume q1 = q2 = q. Using the formula mentioned earlier, we get U1= kq^2/sU2= kq^2/2s. Therefore, the ratio of their potential energies is U2/U1= kq^2/2s / kq^2/sU2/U1= (kq^2/2s) × (s/kq^2)U2/U1= 1/2.

Therefore, the ratio of their potential energies is 1:2.

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in one cycle a heat engine absorbs 480 j from a high-temperature reservoir and expels 320 j to a low-temperature reservoir. if the efficiency of this engine is 56% of the efficiency of a carnot engine, what is the ratio of the low temperature to the high temperature in the carnot engine?

Answers

The ratio of the low temperature to high temperature of the Carnot engine is 2.38.

What is the efficiency of Carnot engine?

The efficiency of the Carnot engine can be defined as the ratio of network done per cycle by the engine to the heat energy absorbed by the engine per cycle by the working substance from the source.

Efficiency = 1 - (Tlow/Thigh)

Heat absorbed by engine = 480J

Heat expelled by engine = 320J

Efficiency of the engine = 56% of efficiency of Carnot engine

The ratio of low temperature to high temperature in the Carnot engine.

Let's assume the efficiency of the Carnot engine is 'ηc' = 1 - T₂/T₁

Where, T₂ = Low temperature and T₁ = High temperature

To calculate the efficiency of the engine given, η = (Q1 - Q2)/Q1

η = (480 - 320)/480

η = 160/480

η = 1/3

η = 33.33%

Now, η = 56% × ηc

0.56ηc = 1/3ηc = (1/3)/0.56 = 0.58

As we already know, ηc = 1 - T₂/T₁

T₂/T₁ = 1 - ηc

T₂/T₁ = 1 - 0.58

T₂/T₁ = 0.42

T₁/T₂ = 1/0.42

T₁/T₂ = 2.38

Therefore, the ratio of low temperature to high temperature in the given Carnot engine with an efficiency of 56% will be about 2.38.

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A student walks 1.0 kilometer due east and 1.0 kilometer due south. Then
she runs 2.0 kilometers due west. The magnitude of the student's
resultant displacement is closestto
A. 3.4 km
B. 1.4 km
C. 4.0 km
D. O km

Answers

The resulting displacement will be 3.4 km. The correct option is A.

How to calculate the displacement?

The displacement is calculated by finding the displacement from east to west, which is 2.0 km, and subtracting the displacement from north to south, which is 1.0 km.

A student walks 1.0 kilometers due east and 1.0 kilometers due south. Then she runs 2.0 kilometers due west. The magnitude of the student's resultant displacement is closest to 3.4 km.

To begin with, we may use the Pythagorean Theorem to determine the resultant displacement's magnitude. The Pythagorean Theorem is a formula that is used to determine the length of a right triangle's sides when one is missing. This theorem is used to calculate the magnitude of the resultant displacement, which is a quantity. It's a good idea to draw a diagram to help you understand the problem.

Here's a rough sketch of the scenario: We will now apply the Pythagorean theorem in this way: The resultant displacement's magnitude is 3.4 kilometers. Thus, the correct option is A.

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Discuss three applications of the effects of surface tension.

Answers

1. Surface tension: is the property of the liquid by virtue of which the free surface of liquid at rest tends to have minimum surface area and as such it behaves as if covered with a stretched membrane.

2. Effect of Surface tension:
Surface tension of soap solution is less, it can spread over large areas and wash clothes more effectively, since the dirt particles stick to the soap molecules.
In soldering, addition of flux reduces the surface tension of molten tin. Hence, it spreads.
Antiseptics like dettol have low surface tension, so that they spread faster.
Surface tension prevents water from passing through the pores of an umbrella.
A duck is able to float on water as its feathers secrete oil that lowers the surface tension of water.

hydroelectric, wind, geothermal, and parabolic solar collection all rely on spinning turbines (connected to a generator) to produce electricity. explain how each provides the force to do so.

Answers

Hydroelectric energy is generated by capturing the energy of flowing water. As water flows through a turbine, the blades of the turbine spin and generate electricity.

How does the different energies provide force?

Wind energy is generated by capturing the kinetic energy of the wind. As wind passes through the turbine, the blades spin and generate electricity.

Geothermal energy is generated by harnessing the natural heat of the Earth’s core. Heat from the Earth’s core is used to generate steam, which is then used to spin a turbine and generate electricity.

Parabolic solar collection is a method of collecting the sun’s energy using large reflective mirrors. The mirrors focus the sunlight onto a central point, which is then used to spin a turbine and generate electricity.

Thus, all of these power sources rely on spinning turbines connected to a generator to produce electricity.

Answers

"The surface of the sun appears sharp in visible light because the photosphere is thin compared to the other layers in the sun."

Most of the electromagnetic energy that reaches the earth begins in the photosphere, the area of the sun that is visible to us. The photosphere is referred to as the sun's surface, despite the fact that it is a gaseous entity.

The gas in the photosphere appears to have a sharp surface, but in reality, it is heavier lower in the Sun and less dense higher up. It is more transparent the less thick it is. The area of the gas that is visible to us is where it has largely become translucent. About 300 km of this layer are deep.

The photosphere is the line separating the core of the Sun from its atmosphere. It is the part of the Sun's surface that is visible to us. The photosphere is not like a planet's surface; even if you could stand in the sun, you couldn't do so on the photosphere.

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