Problem 586
Question
If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be .... \(\\{\mathrm{A}\\}[365 \times 3]\) \(\\{\mathrm{B}\\}[365 \times 27]\) \(\\{\mathrm{C}\\}[365 \times(3 \sqrt{3})]\) \(\\{\mathrm{D}\\}[365 /(3 \sqrt{3})]\)
Step-by-Step Solution
Verified Answer
\(T_2 = 365 \sqrt{27}\)
1Step 1: Understand Kepler's Third Law
Kepler's Third Law states that the square of the time period (T) of a planet's orbit is proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, it is represented as:
\(T^2 \propto a^3\)
For any two orbits, we can write the equation as:
\(\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}\)
Here, \(T_1\) and \(T_2\) represent the time periods of the two orbits, and \(a_1\) and \(a_2\) are their respective semi-major axes.
2Step 2: Set up the equation with the given information
In this case, the first orbit is the present orbit of the Earth with a distance of a, and the second orbit is when the Earth's distance becomes three times the present distance, i.e., 3a. We are given that the number of days in one year currently is 365 days, i.e., \(T_1 = 365\). We have to find the number of days (\(T_2\)) in one year when the Earth's distance is three times the present distance.
Using the relation from Kepler's Third Law:
\(\frac{365^2}{T_2^2} = \frac{a^3}{(3a)^3}\)
3Step 3: Solve for \(T_2\)
Now, we need to solve this equation for \(T_2\). We can start by simplifying the equation and canceling out the terms:
\(\frac{365^2}{T_2^2} = \frac{a^3}{27a^3}\)
From this, we can observe that \(a^3\) will cancel out on both sides:
\(\frac{365^2}{T_2^2} = \frac{1}{27}\)
Next, we need to isolate \(T_2\) on one side of the equation, so we can cross-multiply:
\(T_2^2 = 27 \cdot 365^2\)
To find the value of \(T_2\), we can take the square root of both sides:
\(T_2 = \sqrt{27 \cdot 365^2}\)
Now, we can compute the value of \(T_2\):
\(T_2 = 365 \sqrt{27}\)
4Step 4: Match the answer with the given options
The value of \(T_2\) that we found is \(365 \sqrt{27}\). If we look at the answer choices, we can see that the correct option is:
\({\mathrm{B}}[365 \times 27]\)
Key Concepts
Planetary OrbitsSemi-Major AxisTime Period of OrbitGravitational Laws
Planetary Orbits
Planetary orbits are the paths planets follow around the sun. These orbits are not perfect circles; instead, they are elliptical.
Each orbit varies in size, shape, and orientation, depending on the gravitational forces acting upon the planet.
In our solar system, every planet, including Earth, follows an orbit around the sun due to its gravitational pull.
Each orbit varies in size, shape, and orientation, depending on the gravitational forces acting upon the planet.
In our solar system, every planet, including Earth, follows an orbit around the sun due to its gravitational pull.
- The sun sits at one of the two foci of the elliptical orbit.
- The distance between the sun and the planet changes as it moves through its orbit.
- Orbits are maintained by the gravitational balance between the planet and the sun.
Semi-Major Axis
The semi-major axis is an important dimension of an elliptical orbit. It is half the longest diameter of the ellipse and extends from the center of the ellipse to the furthest point on the orbit's edge.
This measurement helps us understand the scale of the orbit and is pivotal in applying Kepler's Third Law.
This measurement helps us understand the scale of the orbit and is pivotal in applying Kepler's Third Law.
- The semi-major axis determines the average distance of a planet from the sun.
- It influences the length of the time period of the orbit.
- In the case of Earth's orbit, altering the semi-major axis affects the year's length.
Time Period of Orbit
The time period of an orbit is the duration it takes for a planet to complete one full revolution around the sun. For Earth, this is famously known as a year, approximately 365 days.
According to Kepler's Third Law, the time period is connected with the orbit's semi-major axis. When the semi-major axis changes, the time it takes to go around the sun changes too, which in turn affects the length of the year.
According to Kepler's Third Law, the time period is connected with the orbit's semi-major axis. When the semi-major axis changes, the time it takes to go around the sun changes too, which in turn affects the length of the year.
- If the semi-major axis increases, the time period increases.
- This means a planet further from the sun takes longer to complete its orbit.
- The time period for orbits is crucial for understanding celestial mechanics.
Gravitational Laws
Gravitational laws describe how objects in space attract each other. Sir Isaac Newton formulated these laws, which explain how distinct masses, like the sun and Earth, exert a force on each other.
These laws underpin our understanding of planetary motion and are reflected in Kepler's work on planetary orbits.
These laws underpin our understanding of planetary motion and are reflected in Kepler's work on planetary orbits.
- Gravity is the force that keeps planets in orbit around the sun.
- It depends on the mass of the objects and the distance between them.
- Greater mass and proximity result in a stronger gravitational pull.
Other exercises in this chapter
Problem 581
The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of \(1500 \ma
View solution Problem 582
An automobile engine develops \(100 \mathrm{kw}\) when rotating at a speed of 1800 r.p.m. what torque does it deliver? \(\\{\mathrm{A}\\} 350 \mathrm{Nm} \quad\
View solution Problem 587
A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be.... \(\\{\mathrm{A}\\} 15: 14
View solution Problem 588
The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ...... \(\\{\mathrm{A}\\} \sqrt
View solution