A particle of mass m having negative charge q | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an elliptical orbit under a central force,the conservation of energy and angular momentum applies.At the points of closest approach $(r_1)$ an

Vedclass · Accepted Answer

$\sqrt{\frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}}$

Vedclass · Answer

(A) For an elliptical orbit under a central force,the conservation of energy and angular momentum applies.At the points of closest approach $(r_1)$ and farthest distance $(r_2)$,the velocity is perpendicular to the radius vector.Conservation of angular momentum: $L = mv_1r_1 = mv_2r_2 \implies v_1 = \frac{L}{mr_1}$ and $v_2 = \frac{L}{mr_2}$.Conservation of energy: $\frac{1}{2}mv_1^2 - \frac{Qq}{4\pi\varepsilon_0r_1} = \frac{1}{2}mv_2^2 - \frac{Qq}{4\pi\varepsilon_0r_2}$.Substituting $v_1$ and $v_2$: $\frac{L^2}{2m}(\frac{1}{r_1^2} - \frac{1}{r_2^2}) = \frac{Qq}{4\pi\varepsilon_0}(\frac{1}{r_1} - \frac{1}{r_2})$.Using $a^2 - b^2 = (a-b)(a+b)$: $\frac{L^2}{2m} \frac{(r_2-r_1)(r_2+r_1)}{r_1^2r_2^2} = \frac{Qq}{4\pi\varepsilon_0} \frac{(r_2-r_1)}{r_1r_2}$.Solving for $L$: $L^2 = \frac{2mQq}{4\pi\varepsilon_0} \frac{r_1r_2}{r_1+r_2} = \frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}$.Thus,$L = \sqrt{\frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}}$.

$A$ particle of mass $m$ having negative charge $q$ moves along an ellipse around a fixed positive charge $Q$ such that its maximum and minimum distances from the fixed charge are $r_1$ and $r_2$ respectively. The angular momentum $L$ of this particle is:

Similar Questions

The solar constant on the surface of the earth is $S$. What will be its value on the surface of another planet which is about $5.3 \, AU$ away from the sun?

Assume that a tunnel is dug along a chord of the Earth,at a perpendicular distance $(R / 2)$ from the Earth's centre,where $R$ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel,it will execute a simple harmonic motion with a time period:

Two satellites of masses $m$ and $1.5 m$ are revolving around the earth in two circular orbits of heights $R_E$ and $2 R_E$ respectively,where $R_E$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces exerted by the two satellites on the earth is

$A$ star of mass $M$ and radius $R$ is made up of gases. The average gravitational pressure compressing the star due to the gravitational pull of the gases making up the star depends on $R$ as

$A$ particle of mass $m$ moves in a circular orbit under the central potential field,$U(r) = -\frac{C}{r}$,where $C$ is a positive constant. The correct radius-velocity graph of the particle's motion is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module