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If the Earth were suddenly stopped in its orbit, how long would it take for it to collide with the Sun? [Regard the Sun as a fixed point mass. You may make use of the formula for the period of the Earth's orbit.]

Short Answer

Expert verified
The Earth would take approximately \(t\) seconds to collide with the Sun if it were suddenly stopped in its orbit.

Step by step solution

01

Calculate the distance from the Earth to the Sun

Using the formula \(T = 2 \pi \sqrt {\frac {a^3} {GM}}\), we can rearrange for \(a\) (distance from the Earth to the Sun) and substitute the known values for \(T\), \(M\), and \(G\). <: T = 2 \pi \sqrt {\frac {a^3} {GM}} \Rightarrow a = \sqrt[3]{\frac {G M T^2} {4 \pi^2}} :> Let's plug in the values, \(T = 365.25 * 24 * 60 * 60\ sec\), \(G = 6.674 * 10^{-11} \,m^3kg^{-1}s^{-2}\), and \(M = 1.989 * 10^{30}\,kg\). This gives us \(a\).
02

Calculate the time for collision

Next, we can use the equation for the time it takes for an object to fall a distance \(d\) under uniform gravity \(g\): \(t = \sqrt{2d/g}\). We need to bear in mind that the acceleration due to gravity at the surface of the Earth will not be the same as at the distance of the Earth from the Sun. So \(g\) can furthermore be refined to the constant of universal gravitation multiplied by the mass of the Sun divided by the distance squared: \(g = GM/a^2\). Substitution gives us \(t\).
03

Simplify the expression

With the known values calculated and substituted in, simplify the expression to achieve the final answer. It should take approximately \(t\) seconds for the Earth to collide with the Sun if it was suddenly stopped in its orbit.

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Key Concepts

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

Gravitational Attraction
Gravitational attraction is the force that draws two masses toward each other. It is governed by Newton's law of universal gravitation, which states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The formula can be expressed as: \[\begin{equation} F = G \frac{m_1 m_2}{r^2} \end{equation}\]where:
  • F is the gravitational force between the masses,
  • G is the gravitational constant ( 6.674 \times 10^{-11} \ m^3kg^{-1}s^{-2}),
  • m_1 and m_2 are the masses of the two objects, and
  • r is the distance between the centers of the two masses.
In the orbital mechanics problem described, the two masses in question are Earth and the Sun. The gravitational force between them keeps Earth in a stable orbit around the Sun. If Earth were to stop in its orbit, the gravitational attraction would cause the Earth to begin falling directly toward the Sun.
Kepler's Laws of Planetary Motion
Kepler's laws of planetary motion describe how planets orbit around the Sun. These laws are critical to understanding orbital mechanics and solving related problems.
  • First Law (The Law of Ellipses)

    Planets move in elliptical orbits with the Sun at one focus. The path of the Earth around the Sun is not a perfect circle but an ellipse, though it is fairly close to circular.
  • Second Law (The Law of Equal Areas)

    A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This implies that a planet moves faster when it is closer to the Sun and slower when it is farther away.
  • Third Law (The Law of Harmonies)

    The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. This law provides a relation between the time a planet takes to orbit the Sun (its period) and its average distance from the Sun.
In our exercise, the third law is particularly relevant. It allows us to calculate the average distance between the Earth and the Sun, referred to as the semi-major axis ( a), from the period of Earth's orbit, thus setting up the basis for determining the time it would take for the Earth to fall into the Sun if it stopped in its orbit.
Free-fall Time Calculation
The free-fall time calculation refers to determining how long it would take an object to fall a certain distance under the influence of gravity alone. This is an important concept in physics and is particularly relevant to understanding how objects behave under the influence of gravitational forces in space.The time ( t) it takes for an object starting from rest to fall a distance ( d) under constant acceleration ( g) is given by: \[\begin{equation} t = \sqrt{\frac{2d}{g}} \end{equation}\]For the Earth falling towards the Sun, the distance is the average distance from the Earth to the Sun, and the acceleration due to gravity ( g) varies as we move away from the Earth's surface. The acceleration due to gravity at any distance from the Sun can be calculated using the modified gravitational formula: \[\begin{equation} g = \frac{GM}{r^2} \end{equation}\]where ( M) is the mass of the Sun, and ( r) is the distance between the center of the Sun and the Earth. After calculating ( g) and using the average distance as our ( d), we can determine the free-fall time. This exercise illustrates the application of these principles in a hypothetical scenario where the Earth stops its orbit and experiences free fall towards the Sun.

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Most popular questions from this chapter

It is required to project a body from a point on level ground in such a way as to clear a thin vertical barrier of height \(h\) placed at distance \(a\) from the point of projection. Show that the body will just skim the top of the barrier if $$ \left(\frac{g a^{2}}{2 u^{2}}\right) \tan ^{2} \alpha-a \tan \alpha+\left(\frac{g a^{2}}{2 u^{2}}+h\right)=0 $$ where \(u\) is the speed of projection and \(\alpha\) is the angle of projection above the horizontal. Deduce that, if the above trajectory is to exist for some \(\alpha\), then \(u\) must satisfy $$ u^{4}-2 g h u^{2}-g^{2} a^{2} \geq 0 $$

A body is projected vertically upwards with speed \(u\) and moves under uniform gravity in a medium that exerts a resistance force proportional to the fourth power its speed and in which the body's terminal speed is \(V\). Find the maximum height above the starting point attained by the body. Deduce that, however large \(u\) may be, this maximum height is always less than \(\pi V^{2} / 4 g\).

A mortar gun, with a maximum range of \(40 \mathrm{~m}\) on level ground, is placed on the edge of a vertical cliff of height \(20 \mathrm{~m}\) overlooking a horizontal plain. Show that the horizontal range \(R\) of the mortar gun is given by $$R=40\left\\{\sin \alpha+\left(1+\sin ^{2} \alpha\right)^{\frac{1}{2}}\right\\} \cos \alpha$$ where \(\alpha\) is the angle of elevation of the mortar above the horizontal. [Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).] Evaluate \(R\) (to the nearest metre) when \(\alpha=45^{\circ}\) and \(35^{\circ}\) and confirm that \(\alpha=45^{\circ}\) does not yield the maximum range. [Do not try to find the optimum projection angle this way. See Problem \(4.22\) below. \(]\)

A particle of mass \(m\) is attached to the highest point of a smooth rigid sphere of radius \(a\) by a light inextensible string of length \(\pi a / 4\). The particle moves in contact with the outer surface of the sphere, with the string taut, and describes a horizontal circle with constant speed \(u\). Find the reaction of the sphere on the particle and the tension in the string. Deduce the maximum value of \(u\) for which such a motion could take place. What will happen if \(u\) exceeds this value?

A body has terminal speed \(V\) when falling in still air. What is its terminal velocity (relative to the ground) when falling in a steady horizontal wind with speed \(U\) ?

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