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Distance between the centres of two stars is \(10 a\). The masses of these stars are \(M\) and \(16 M\) and their radii \(a\) and \(2 a\) respectively. A body of mass \(m\) is fired straight from the surface of the larger star towards the smaller star. The minimum initial speed for the body to reach the surface of smaller star is (a) \(\frac{2}{3} \sqrt{\frac{G M}{a}}\) (b) \(\frac{3}{2} \sqrt{\frac{5 G M}{a}}\) (c) \(\frac{2}{3} \sqrt{\frac{5 G M}{a}}\) (d) \(\frac{3}{2} \sqrt{\frac{G M}{a}}\)

Short Answer

Expert verified
Option (b): \(\frac{3}{2} \sqrt{\frac{5 G M}{a}}\).

Step by step solution

01

Initial Conditions and Energy Consideration

The problem involves determining the minimum initial velocity of a body so that it can move from the surface of the larger star to that of the smaller star. The key here is to consider energy conservation. Let's start by setting the initial and final total mechanical energies equal. The body starts with initial kinetic energy and potential energy on the surface of the larger star, moving to the potential energy at the surface of the smaller star. Since the speed must be minimal, the final kinetic energy at the surface of the smaller star is zero.
02

Initial Kinetic and Potential Energy

The initial total energy when the body is at the surface of the larger star can be written as: \[E_i = rac{1}{2} m v^2 - rac{G (16M) m}{2a},\]where \(v\) is the initial velocity, and the potential energy is calculated due to the larger star's mass.
03

Reaching the Smaller Star's Surface

When the body reaches the smaller star's surface, its total energy will be entirely potential if it just makes it to the surface (minimal initial speed). Thus,\[E_f = - rac{G M m}{a} - rac{G (16M) m}{10 a}.\]This is the sum of the gravitational potentials due to each star at the location of the smaller star's surface.
04

Applying Conservation of Energy

By energy conservation principle, we have:\[E_i = E_f.\] Substitute in from previous steps to get:\[\frac{1}{2} m v^2 - \frac{G (16M) m}{2a} = - \frac{G M m}{a} - \frac{G (16M) m}{10 a}.\]
05

Solve for Initial Velocity

Simplify the equation:\[\frac{1}{2} m v^2 = \frac{G (16M) m}{2a} - \left(-\frac{G M m}{a} - \frac{G (16M) m}{10 a}\right).\] Calculate the terms on the right:\[\frac{G (16M) m}{2a} + \frac{G M m}{a} + \frac{G (16M) m}{10 a}.\] The terms combine to give:\[\frac{1}{2} m v^2 = \frac{45 G M m}{10 a}.\]Thus,\[v^2 = \frac{9 G M}{a}.\]Finally, solve for \(v\):\[v = \frac{3}{2} \sqrt{\frac{5 G M}{a}}.\]
06

Conclusion

From the calculations above, the minimum initial speed required is \(v = \frac{3}{2} \sqrt{\frac{5 G M}{a}}\). This matches option (b).

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Key Concepts

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

Conservation of Energy
In physics, the conservation of energy is a fundamental concept that states that the total energy of an isolated system remains constant. For our problem, this principle is crucial because it allows us to determine the minimum initial speed required for the mass to travel from the larger star to the smaller star.

The body we are analyzing starts with a certain amount of kinetic and potential energy. As it moves between the two stars, its kinetic energy changes, but the sum of kinetic and potential energy – its total mechanical energy – remains constant. This means we can set the initial total energy equal to the final total energy. By doing so, we can solve for the unknown initial velocity. This concept is essential to solving two-body problems where forces, such as gravity, do work that can change the forms of energy but not the total amount.
Gravitational Potential Energy
Gravitational potential energy is the energy held by an object because of its position relative to a massive body, like a star or planet. In simpler terms, it's the energy an object has due to the gravitational pull exerted on it.

In our exercise, the body has potential energy because of its position in the gravitational fields of both stars. At the surface of the larger star, it starts with a high potential energy which decreases as it moves toward the smaller star. The formula for gravitational potential energy between two masses, say a mass "m" and the larger star mass "16M", is given by:\[- \frac{G(16M)m}{2a} \]where:
  • \(G\) is the gravitational constant
  • \(M\) and \(16M\) are the masses of the stars
  • \(a\) and \(2a\) are the radii of the stars
Gravitational potential energy within the vicinity of these masses plays a pivotal role in determining the initial conditions for the body's motion.
Two-body Problem
The two-body problem involves calculating the motion of two masses that interact with each other solely through mutual forces, such as gravity. In our exercise, the two stars and the mass "m" illustrate this well-known physics problem.

Here, we simplify the model by treating other forces as negligible so that the gravitational interaction dominates. The gravitational influence of the two stars controls the trajectory and energy states of the body being fired. By considering only these two celestial bodies and neglecting any other forces or influences, our calculation becomes more manageable and focused.

The initial conditions set on the surface of the larger star and the aim to reach the smaller star's surface simplify the computations, and conservation of energy makes it possible to find a clean analytical solution to this complex interaction.
Escape Velocity
Escape velocity is the minimum velocity an object must have to break free from the gravitational influence of a celestial body without further propulsion. It’s a concept closely tied to gravitational potential energy and kinetic energy.

When we fire a body from the surface of the larger star, the minimum speed needed to "escape" or at least reach the surface of the smaller star is akin to calculating an escape velocity. Therefore, our task is determining this minimum initial speed.

By formulating the problem using the principle of conservation of energy, we ensure that energy is just sufficient to reach the desired point with zero kinetic energy at the end, because any excess would mean a higher initial speed. Hence, our calculated speed resembles escape velocity calculations:\[v = \frac{3}{2} \sqrt{\frac{5 G M}{a}}\]
This specific speed allows the body "m" to reach the other star, mirroring the idea of escaping a gravitational field but targeted toward a specific endpoint.

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Most popular questions from this chapter

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is $$ I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be $$ I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right) $$ When \(L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}\)

Angular momentum in absence of any external torque remains constant. If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e., \(\tau=0\) $$ \begin{aligned} &\therefore \quad \frac{d L}{d t}=0 \\ &\Rightarrow \quad l_{1} \omega_{1}=I_{2} \omega_{2} \\ &\text { Here, } M=2 \mathrm{~kg}, m=0.25 \mathrm{~kg}, r=0.2 \mathrm{~m} \\ &\omega_{1}=30 \mathrm{rad} \mathrm{s}^{-1} \end{aligned} $$ Hence, we get after putting the given values in Eq. (i) $$\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25)(0.2)^{2} \times \omega_{2}$$ $$\begin{array}{ll}\Rightarrow & 1.2=0.05 \omega_{2} \\ \therefore & \omega_{2}=24 \mathrm{rad} \mathrm{s}^{-1}\end{array}$$

Three particles each of mass \(m\) rotate in a circle of radius \(r\) with uniform angular speed \(\omega\) under their mutual gravitational attraction. If at any instant the points are on the vertex of an equilateral of side \(L\), then angular velocity \(\omega\) is (a) \(\sqrt{\frac{2 G m}{L^{3}}}\) (b) \(\sqrt{\frac{3 G m}{L^{3}}}\) (c) \(\sqrt{\frac{5 G m}{L^{3}}}\) (d) \(\sqrt{\frac{G m}{L^{3}}}\)

A particle of mass \(m\) is placed at the centre of a uniform spherical shell of mass \(3 \mathrm{~m}\) and radius \(R\). The gravitational potential on the surface of the shell is (a) \(-\frac{G m}{R}\) (b) \(-\frac{3 \mathrm{Gm}}{R}\) (c) \(-\frac{4 G m}{R}\) (d) \(-\frac{2 G m}{R}\)

For a body lying on the equator to appear weightless, what should be the angular speed of the earth? (Take \(g=10 \mathrm{~ms}^{-2}\); radius of earth \(=6400 \mathrm{~km}\) ) (a) \(0.125 \mathrm{rads}^{-1}\) (b) \(1.25 \mathrm{rads}^{-1}\) (c) \(1.25 \times 10^{-3} \mathrm{rads}^{-1}\) (d) \(1.25 \times 10^{-2} \mathrm{rads}^{-1}\)

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