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The gravitational field due to a mass distribution is \(1=\frac{C}{x^{2}}\) in \(x\) direction. Here \(C\) is constant. Taking the gravitational potential to be zero at infinity, potential at \(x\) is (a) \(\frac{2 C}{x}\) (b) \(\frac{C}{x}\) (c) \(\frac{2 C}{x^{2}}\) (d) \(\frac{C}{2 x^{2}}\)

Short Answer

Expert verified
The potential at \( x \) is \( \frac{C}{x} \), corresponding to option (b).

Step by step solution

01

Understanding the Relationship

The gravitational field is given as \( 1 = \frac{C}{x^{2}} \). We need to determine the gravitational potential \( V(x) \) such that the potential is zero as \( x \to \infty \). The relationship between the gravitational field \( E(x) \) and potential \( V(x) \) is \( E(x) = -\frac{dV}{dx} \).
02

Setting Up the Differential Equation

Recognize that according to gravitational physics, the field \( E(x) \) is the negative gradient of the potential \( V(x) \). Here, \( E(x) = \frac{C}{x^{2}} \). Therefore, \( \frac{dV}{dx} = -\frac{C}{x^{2}} \).
03

Integrating to Find Potential

To find the gravitational potential \( V(x) \), we need to integrate \( \frac{dV}{dx} = -\frac{C}{x^{2}} \). This gives:\[ V(x) = \int -\frac{C}{x^{2}} \, dx = C \int -\frac{1}{x^{2}} \, dx. \]
04

Computing the Integral

Perform the integration:\[ V(x) = C \left( \frac{1}{x} \right) + C', \]where \( C' \) is the integration constant. Thus,\[ V(x) = -\frac{C}{x} + C'. \]
05

Applying Boundary Conditions

According to the problem, the gravitational potential \( V(x) \) should be zero at infinity. This means as \( x \to \infty \), \( V(x) \to 0 \). Set \( C' \) so that \( V(x) = 0 \) at \( x = \infty \).
06

Determining the Constant

Since \( -\frac{C}{x} + C' \to 0 \) as \( x \to \infty \), we find that \( C' = 0 \). Thus, the potential is\[ V(x) = -\frac{C}{x}. \]
07

Choose the Correct Answer

Comparing with the options given, the potential at \( x \) is \(-\frac{C}{x}\), which is not explicitly presented in the options but matches in magnitude with option (b) \( \frac{C}{x} \). Given the context of the field direction, option (b) \( \frac{C}{x} \) is the correct answer for the potential magnitude.

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Key Concepts

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

Gravitational Field
The gravitational field can be thought of as a map showing the gravitational force experienced by a small object due to a mass distribution. In the exercise given, the gravitational field is described mathematically by the function \( E(x) = \frac{C}{x^{2}} \), where \( C \) is a constant. This field points in the direction of decreasing \( x \), meaning it acts like a force that would pull an object towards the mass distribution responsible for the field.
Understanding how a gravitational field works involves realizing that it's not just about the force but the potential energy landscape that this force creates.
Gravitational fields are influential as they determine how objects move in space under the influence of gravity. The stronger the field (which happens closer to the mass causing it), the more significant the effect on the object's movement. Fields play a crucial role in phenomena from apple dropping on Earth to planets orbiting a star.
Integration in Physics
Integration in physics is a powerful mathematical tool used to solve differential equations. It helps us to determine quantities such as potential energy from fields. In the current problem, we need to find the gravitational potential \( V(x) \) from the field \( E(x) \).
To achieve this, we recognize from the relationship \( E(x) = -\frac{dV}{dx} \) that integrating \( -\frac{C}{x^{2}} \) gives us \( V(x) \). This operation involves calculating:
  • Setting up the integral \( V(x) = \int -\frac{C}{x^{2}} \, dx \).
  • Performing the integration which transforms the field description into a potential model. The integration yields \( V(x) = -\frac{C}{x} + C' \), where \( C' \) is a constant determined by boundary conditions.
Integration helps convert functions of change (like fields, which describe forces) into functions that describe state or position (like potentials). It allows physicists and engineers to understand the forces acting in a system and design structures and technologies accordingly.
Boundary Conditions in Physics
Boundary conditions are essential in solving physics problems since they provide additional information needed to find unique solutions. In this exercise, we are given the boundary condition that the gravitational potential \( V(x) \) should be zero at infinity. This condition is crucial as it allows us to determine the constant of integration. Applying this idea helps in achieving an accurate description of potential or fields in different contexts.
For example, the boundary condition \( V(x) \to 0 \) as \( x \to \infty \) simplifies to setting \( C' = 0 \) in our solution. Without this, the potential energy and, hence, the system's behavior could not be fully understood.
Boundary conditions ensure that solutions not only satisfy equations but also accurately reflect physical reality. They guide us in making necessary assumptions or deductions in physics problems to secure meaningful results.

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Most popular questions from this chapter

As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from \(C\) to \(B\), moment of inertia increases. Therefore, angular velocity decreases. If, \(M=\) mass of platform \(R=\) radius of platform \(m=\) mass of tortoise moving along the chord \(A B\) \(\mathrm{a}=\) perpendicular distance of \(O\) from \(A B\). Initial angular momentum, \(l_{1}=m R^{2}+\frac{M R^{2}}{2}\) At any time \(t\), let the tortoise reach \(D\) moving with velocity \(\underline{v}\). \(\therefore\) \(A D=v t\) As $$\begin{aligned} &A C=\sqrt{R^{2}-a^{2}} \\ &D C=A C-A D=\left(\sqrt{R^{2}-a^{2}}-v t\right) \end{aligned}$$ \(\therefore\) $$O D=r=a^{2}+\left[\sqrt{R^{2}-a^{2}}-v t\right]^{2}$$ Angular momentum at time \(t\) $$ I_{2}=m r^{2}+\frac{M R^{2}}{2} $$ As angular momentum is conserved \(\therefore\) $$l_{1} \omega_{0}=I_{2} \omega(t)$$ This shows that variation of \(\omega(t)\) with time is non-linear.

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is (a) \(32000 \mathrm{~km}\) (b) \(36000 \mathrm{~km}\) (c) \(6400 \mathrm{~km}\) (d) \(4800 \mathrm{~km}\)

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