Pravegaa PYQ Series · Volume 2: Classical Mechanics & Special Relativity

CM04: Central Forces and Gravitation

78 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

IIT JAM: 21
GATE: 4
CSIR NET: 15
JEST: 14
TIFR GS: 24

CM04.1IIT JAM 2026, Q52NAT | 2 marks

A spacecraft is placed $200\,\text{km}$ above Earth in a circular orbit. The minimum change in the speed required to place the spacecraft in a parabolic orbit, in km/s, is . (Rounded off to one decimal place)

[Use $G=6.67\times10^{-11}\,\text{Nm}^2/\text{kg}^2$, mass of Earth $=6\times10^{24}\,\text{kg}$, radius of Earth $=6400\,\text{km}$]

Show solution

The orbit radius is $r=R_E+h=6400+200=6600$ km. The circular orbital speed is \[ v_c=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{6.6\times10^{6}}}=7787\ \text{m/s}. \] A parabolic orbit has zero total energy, so it needs the escape speed $v_e=\sqrt{2GM/r}=\sqrt2\,v_c$. The minimum boost (applied tangentially) is \[ \Delta v=(\sqrt2-1)v_c=0.4142\times7787=3225\ \text{m/s} \] \[ \boxed{\Delta v\approx3.2\ \text{km/s}} \]

Answer: $3.2$ (range 3.0 to 3.5)

CM04.2IIT JAM 2025, Q16MCQ | 2 marks

Three particles of equal mass $M$, interacting via gravity, lie on the vertices of an equilateral triangle of side $d$, as shown in the figure below. The whole system is rotating with an angular velocity $\omega$ about an axis perpendicular to the plane of the system and passing through the center of mass. The value of $\omega$, for which the distance between the masses remains $d$, is ($G$ is the universal gravitational constant)

$\sqrt{\dfrac{2GM}{d^3}}$
$\sqrt{\dfrac{3GM}{d^3}}$
$\sqrt{\dfrac{GM}{3d^3}}$
$\sqrt{\dfrac{GM}{d^3}}$

Show solution

Each mass moves on a circle of radius $r=d/\sqrt3$ about the centroid. The two attractions on a mass, each $GM^2/d^2$, make $30^\circ$ with the line to the centroid, so the net inward force is \[F=2\,\frac{GM^2}{d^2}\cos30^\circ=\frac{\sqrt3\,GM^2}{d^2}.\] Equating this to the centripetal force, \[M\omega^2\frac{d}{\sqrt3}=\frac{\sqrt3\,GM^2}{d^2}\;\Rightarrow\;\boxed{\omega=\sqrt{\dfrac{3GM}{d^3}}}\]

Answer: (B) $\sqrt{3GM/d^3}$

CM04.3IIT JAM 2025, Q32MSQ | 2 marks

Two particles of masses $m_1$ and $m_2$, interacting via gravity, rotate in circular orbits about their common center of mass with the same angular velocity $\omega$. For masses $m_1$ and $m_2$, respectively,

$r_1$ and $r_2$ are the constant distances from the center of mass,
$L_1$ and $L_2$ are the magnitudes of the angular momenta about the center of mass, and
$K_1$ and $K_2$ are the kinetic energies.

Which of the following is(are) correct?

($G$ is the universal gravitational constant)

$\dfrac{L_1}{L_2}=\dfrac{m_2}{m_1}$
$\dfrac{K_1}{K_2}=\dfrac{m_2}{m_1}$
$\omega=\sqrt{\dfrac{G(m_1+m_2)}{(r_1+r_2)^3}}$
$m_2r_1=m_1r_2$

Show solution

With $r_0=r_1+r_2$, the centre-of-mass condition $m_1r_1=m_2r_2$ gives \[r_1=\frac{m_2r_0}{m_1+m_2},\qquad r_2=\frac{m_1r_0}{m_1+m_2}.\] Option (D), $m_2r_1=m_1r_2$, would need $r_1/r_2=m_1/m_2$, the inverse of the correct ratio, so it is wrong.

(A) $\dfrac{L_1}{L_2}=\dfrac{m_1r_1^2\omega}{m_2r_2^2\omega}=\dfrac{(m_1r_1)r_1}{(m_2r_2)r_2}=\dfrac{r_1}{r_2}=\dfrac{m_2}{m_1}$. Correct.

(B) $\dfrac{K_1}{K_2}=\dfrac{\frac12m_1r_1^2\omega^2}{\frac12m_2r_2^2\omega^2}=\dfrac{r_1}{r_2}=\dfrac{m_2}{m_1}$. Correct.

(C) For $m_1$, gravity provides the centripetal force: \[\frac{Gm_1m_2}{r_0^2}=m_1\omega^2r_1=\frac{m_1m_2}{m_1+m_2}\,\omega^2r_0\;\Rightarrow\;\omega=\sqrt{\frac{G(m_1+m_2)}{(r_1+r_2)^3}} .\] Correct. Hence $\boxed{\text{(A), (B), (C)}}$.

Answer: (A), (B), (C)

CM04.4IIT JAM 2025, Q56NAT | 2 marks

A planet rotates in an elliptical orbit with a star situated at one of the foci. The distance from the center of the ellipse to any foci is half of the semi-major axis. The ratio of the speed of the planet when it is nearest (perihelion) to the star to that at the farthest (aphelion) is . (in integer)

Show solution

With semi-major axis $a$, the focal distance is $ae=a/2$, so \[r_{\min}=a-\frac a2=\frac a2,\qquad r_{\max}=a+\frac a2=\frac{3a}{2}.\] At both apsides the velocity is perpendicular to the radius, so conservation of angular momentum gives \[mv_pr_{\min}=mv_ar_{\max}\;\Rightarrow\;\frac{v_p}{v_a}=\frac{r_{\max}}{r_{\min}}=\frac{3a/2}{a/2}=\boxed{3}\]

Answer: 3

CM04.5IIT JAM 2024, Q35MSQ | 2 marks

A particle of mass $m$, having an energy $E$ and angular momentum $L$, is in a parabolic trajectory around a planet of mass $M$. If the distance of the closest approach to the planet is $r_{\mathrm m}$, which of the following statement(s) is(are) true?

($G$ is the Gravitational constant)

$E>0$
$E=0$
$L=\sqrt{2GMm^2r_{\mathrm m}}$
$L=\sqrt{2GM^2mr_{\mathrm m}}$

Show solution

For a Kepler orbit, $E<0$ gives an ellipse, $E=0$ a parabola and $E>0$ a hyperbola. So for a parabolic trajectory $E=0$: (B) is true and (A) is false.

The energy is \[ E=\frac12m\dot r^2+\frac{L^2}{2mr^2}-\frac{GMm}{r}. \] At the closest approach $r=r_{\mathrm m}$, $\dot r=0$: \[ 0=\frac{L^2}{2mr_{\mathrm m}^2}-\frac{GMm}{r_{\mathrm m}}\quad\Rightarrow\quad L^2=2GMm^2r_{\mathrm m},\qquad L=\boxed{\sqrt{2GMm^2r_{\mathrm m}}} \] So (C) is true and (D) is false.

Answer: (B), (C)

CM04.6IIT JAM 2024, Q55NAT | 2 marks

A satellite of mass $10\ \mathrm{kg}$, in a circular orbit around a planet, is having a speed $v=200\ \mathrm{m/s}$. The total energy of the satellite is kJ. (Rounded off to nearest integer)

Show solution

For a circular orbit the gravitational force supplies the centripetal force: \[ \frac{GMm}{r^2}=\frac{mv^2}{r}\quad\Rightarrow\quad\frac{GMm}{r}=mv^2 . \] \[ E=\frac12mv^2-\frac{GMm}{r}=\frac12mv^2-mv^2=-\frac12mv^2=-\frac12\times10\times(200)^2\ \mathrm J=-2\times10^{5}\ \mathrm J=\boxed{-200\ \mathrm{kJ}} \]

Answer: $-200$ (range $-200$ to $-200$)

CM04.7IIT JAM 2023, Q40MSQ | 2 marks

For a particle moving in a general central force field, which of the following statement(s) is/are true?

The angular momentum is a constant of motion
Kepler's second law is valid
The motion is confined to a plane
Kepler's third law is valid

Show solution

For $\vec F=F(r)\hat r$, the torque about the force centre is $\vec\tau=\vec r\times\vec F=0$, so $\vec L$ is conserved and (A) is true.

Since $\vec L=\vec r\times\vec p$ is fixed, $\vec r$ always lies in the plane perpendicular to $\vec L$, so the motion is planar and (C) is true.

The areal velocity is $\dfrac{dA}{dt}=\dfrac{L}{2m}=$ constant for any central force, which is Kepler's second law; (B) is true.

Kepler's third law, $T^2\propto a^3$, holds only for the inverse-square force ($V\propto-1/r$), not for a general central force. (D) is false. \[\boxed{\text{(A), (B), (C)}}\]

Answer: (A), (B), (C)

CM04.8IIT JAM 2022, Q7MCQ | 1 mark

Similar: CM04.29 (CSIR NET Dec 2023)

A particle of mass $m$ and angular momentum $L$ moves in space where its potential energy is \[U(r)=kr^2\;(k>0)\text{ and $r$ is the radial coordinate.}\] If the particle moves in a circular orbit, then the radius of the orbit is

$\left(\dfrac{L^2}{mk}\right)^{1/4}$
$\left(\dfrac{L^2}{2mk}\right)^{1/4}$
$\left(\dfrac{2L^2}{mk}\right)^{1/4}$
$\left(\dfrac{4L^2}{mk}\right)^{1/4}$

Show solution

The effective potential is $U_{\text{eff}}=\dfrac{L^2}{2mr^2}+kr^2$. A circular orbit sits at its minimum: \[ \left.\frac{dU_{\text{eff}}}{dr}\right|_{r_0}=-\frac{L^2}{mr_0^3}+2kr_0=0\;\Rightarrow\; r_0^4=\frac{L^2}{2mk} \;\Rightarrow\;\boxed{r_0=\left(\frac{L^2}{2mk}\right)^{1/4}} \]

Answer: (B)

CM04.9IIT JAM 2022, Q18MCQ | 2 marks

A planet of mass $m$ moves in an elliptical orbit. Its maximum and minimum distances from the Sun are $R$ and $r$, respectively. Let $G$ denote the universal gravitational constant, and $M$ the mass of the Sun. Assuming $M\gg m$, the angular momentum of the planet with respect to the center of the Sun is

$m\sqrt{\dfrac{2GMRr}{(R+r)}}$
$m\sqrt{\dfrac{GMRr}{2(R+r)}}$
$m\sqrt{\dfrac{GMRr}{(R+r)}}$
$2m\sqrt{\dfrac{2GMRr}{(R+r)}}$

Show solution

Let $J$ be the angular momentum. The energy at distance $\rho$ is $E=\dfrac m2\dot\rho^2+\dfrac{J^2}{2m\rho^2}-\dfrac{GMm}{\rho}$. At the apsides $\dot\rho=0$, and for an ellipse $E=-\dfrac{GMm}{2a}=-\dfrac{GMm}{R+r}$ (since $2a=R+r$). At $\rho=R$: \[ -\frac{GMm}{R+r}=\frac{J^2}{2mR^2}-\frac{GMm}{R} \;\Rightarrow\;\frac{J^2}{2mR^2}=GMm\,\frac{r}{R(R+r)} \;\Rightarrow\;\boxed{J=m\sqrt{\frac{2GMRr}{R+r}}} \]

Answer: (A)

CM04.10IIT JAM 2022, Q44NAT | 1 mark

A satellite is revolving around the Earth in a closed orbit. The height of the satellite above Earth's surface at perigee and apogee are 2500 km and 4500 km, respectively. Consider the radius of the Earth to be 6500 km. The eccentricity of the satellite's orbit is (Round off to 1 decimal place).

Show solution

Distances are measured from the Earth's centre (the focus): \[ r_{\min}=6500+2500=9000\text{ km},\qquad r_{\max}=6500+4500=11000\text{ km}. \] For an ellipse $r_{\min}=a(1-e)$ and $r_{\max}=a(1+e)$, so \[ e=\frac{r_{\max}-r_{\min}}{r_{\max}+r_{\min}}=\frac{2000}{20000}=\boxed{0.1} \]

Answer: 0.1

CM04.11IIT JAM 2021, Q3MCQ | 1 mark

A planet is in a highly eccentric orbit about a star. The distance of its closest approach is $300$ times smaller than its farthest distance from the star. If the corresponding speeds are $v_c$ and $v_f$, then $\dfrac{v_c}{v_f}$ is

$\dfrac{1}{300}$
$\dfrac{1}{\sqrt{300}}$
$\sqrt{300}$
$300$

Show solution

At the closest and farthest points the velocity is perpendicular to the radius vector, so conservation of angular momentum gives \[ m v_c r_c = m v_f r_f \;\Rightarrow\; \frac{v_c}{v_f}=\frac{r_f}{r_c}=\frac{300\,r_c}{r_c} \] \[ \boxed{\frac{v_c}{v_f}=300} \]

Answer: (D) $300$

CM04.12IIT JAM 2020, Q7MCQ | 1 mark

For a particle moving in a central potential, which one of the following statements is correct?

The motion is restricted to a plane due to the conservation of angular momentum
The motion is restricted to a plane due to the conservation of energy only
The motion is restricted to a plane due to the conservation of linear momentum
The motion is not restricted to a plane

Show solution

A central force produces no torque about the centre, so $\vec J=\vec r\times\vec p$ is constant. Since $\vec r\cdot\vec J=0$ at all times, $\vec r$ always lies in the fixed plane perpendicular to $\vec J$. \[ \boxed{\text{Planar, due to conservation of angular momentum}} \]

Answer: (A)

CM04.13IIT JAM 2020, Q16MCQ | 2 marks

Two planets $P_1$ and $P_2$ having masses $M_1$ and $M_2$ revolve around the Sun in elliptical orbits, with time periods $T_1$ and $T_2$ respectively. The minimum and maximum distances of planet $P_1$ from the Sun are $R$ and $3R$ respectively, whereas for planet $P_2$ these are $2R$ and $4R$, respectively, where $R$ is a constant. Assuming $M_1$ and $M_2$ are much smaller than the mass of the Sun, the magnitude of $\dfrac{T_2}{T_1}$ is

$\dfrac23\sqrt{\dfrac{2M_1}{3M_2}}$
$\dfrac32\sqrt{\dfrac{3M_2}{2M_1}}$
$\dfrac32\sqrt{\dfrac32}$
$\dfrac23\sqrt{\dfrac23}$

Show solution

By Kepler's third law, $T^2=\dfrac{4\pi^2}{GM_s}a^3$, independent of the planet's mass, where the semi-major axis is $a=\frac12(r_{\min}+r_{\max})$: \[ 2a_1=R+3R\Rightarrow a_1=2R,\qquad 2a_2=2R+4R\Rightarrow a_2=3R \] \[ \frac{T_2}{T_1}=\left(\frac{a_2}{a_1}\right)^{3/2}=\left(\frac32\right)^{3/2}=\boxed{\frac32\sqrt{\frac32}} \]

Answer: (C) $\dfrac32\sqrt{\dfrac32}$

CM04.14IIT JAM 2019, Q39MSQ | 2 marks

Consider a classical particle subjected to an attractive inverse-square force field. The total energy of the particle is $E$ and the eccentricity is $\varepsilon$. The particle will follow a parabolic orbit if

$E>0$ and $\varepsilon=1$
$E<0$ and $\varepsilon<1$
$E=0$ and $\varepsilon=1$
$E<0$ and $\varepsilon=1$

Show solution

For the force $F=-k/r^2$ the orbit is a conic with eccentricity \[ \varepsilon=\sqrt{1+\frac{2EJ^2}{mk^2}}, \] where $J$ is the angular momentum. A parabola has $\varepsilon=1$, which requires $E=0$ ($E<0$: ellipse, $\varepsilon<1$; $E>0$: hyperbola, $\varepsilon>1$). \[ \boxed{E=0,\ \varepsilon=1} \]

Answer: (C)

CM04.15IIT JAM 2018, Q3MCQ | 1 mark

There are three planets in circular orbits around a star at distances $a$, $4a$ and $9a$, respectively. At time $t=t_0$, the star and the three planets are in a straight line. The period of revolution of the closest planet is $T$. How long after $t_0$ will they again be in the same straight line?

$8T$
$27T$
$216T$
$512T$

Show solution

By Kepler's third law, $T\propto r^{3/2}$: \[ T_1=k\,a^{3/2}=T,\qquad T_2=k(4a)^{3/2}=8T,\qquad T_3=k(9a)^{3/2}=27T . \] All three planets return to their initial configuration after a time that is a common multiple of the three periods. The least such time is the LCM of $T$, $8T$ and $27T$: \[ t=\mathrm{LCM}(1,8,27)\,T=\boxed{216\,T} \]

Answer: (C) $216T$

CM04.16IIT JAM 2018, Q43NAT | 1 mark

A planet has average density same as that of the earth but it has only $\dfrac18$ of the mass of the Earth. If the acceleration due to gravity at the surface is $g_p$ and $g_e$ for the planet and Earth, respectively, then $\dfrac{g_p}{g_e}=$ .

(Specific your answer upto one digit after the decimal point.)

Show solution

\[ g=\frac{GM}{R^2},\qquad M=\frac43\pi R^3\rho\;\Rightarrow\;R\propto\left(\frac{M}{\rho}\right)^{1/3}. \] For the same density, $g\propto\dfrac{M}{M^{2/3}}=M^{1/3}$, so \[ \frac{g_p}{g_e}=\left(\frac{M_p}{M_e}\right)^{1/3}=\left(\frac18\right)^{1/3}=\boxed{0.5} \]

Answer: 0.5 (range 0.4 to 0.6)

CM04.17IIT JAM 2018, Q52NAT | 2 marks

A body of mass $1$ kg is moving under a central force in an elliptic orbit with semi major axis $1000$ m and semi minor axis $100$ m. The orbital angular momentum of the body is $100\ \text{kg}\,\text{m}^2\text{s}^{-1}$. The time period of motion of the body is hours.

(Specific your answer upto two digits after the decimal point)

Show solution

For a central force the areal velocity is constant: \[ \frac{dA}{dt}=\frac{L}{2m}\;\Rightarrow\;T=\frac{\pi ab}{L/2m}=\frac{2m\,\pi ab}{L}. \] \[ T=\frac{2\times1\times\pi\times1000\times100}{100}\ \text{s}=6283\ \text{s}=\frac{6283}{3600}\ \text{h}=\boxed{1.75\ \text{h}} \]

Answer: 1.75 (range 1.70 to 1.80)

CM04.18IIT JAM 2018, Q53NAT | 2 marks

The moon moves around the earth in a circular orbit with a period of $27$ days. The radius of the earth $(R)$ is $6.4\times10^6$ m and the acceleration due to gravity on the earth surface is $9.8\ \text{ms}^{-2}$. If $D$ is the distance of the moon from the center of the earth, the value of $\dfrac DR$ will be .

(Specific your answer upto two digits after the decimal point)

Show solution

For the circular orbit, with $GM=gR^2$ and $\omega=2\pi/T$, \[ m\omega^2D=\frac{GMm}{D^2}\;\Rightarrow\;D^3=\frac{GM\,T^2}{4\pi^2}=\frac{gR^2T^2}{4\pi^2}\;\Rightarrow\;\left(\frac DR\right)^3=\frac{gT^2}{4\pi^2R}. \] With $T=27\times24\times3600$ s $=2.333\times10^6$ s, \[ \left(\frac DR\right)^3=\frac{9.8\times\left(2.333\times10^6\right)^2}{4\pi^2\times6.4\times10^6}=\frac{5.333\times10^{13}}{2.527\times10^{8}}=2.111\times10^{5} \;\Rightarrow\;\boxed{\frac DR=59.54} \]

Answer: 59.54 (range 59.0 to 60.0)

CM04.19IIT JAM 2017, Q38MSQ | 2 marks

An object of mass $m$ with non-zero angular momentum $\ell$ is moving under the influence of gravitational force of a much larger mass (ignore drag). Which of the following statement(s) is (are) correct?

(A) If the total energy of the system is negative, then the orbit is always circular

(B) The motion of $m$ always occurs in a two-dimensional plane

(D) If the area of the particle's bound orbit is $S$, then its time period is $2mS/\ell$

Show solution

(A) Negative energy gives a bound orbit, which is in general an ellipse; it is circular only for the particular energy $E=-\dfrac{GMm}{2r}$ at the minimum of the effective potential. Incorrect.

(B) The force is central, so $\vec\ell=\vec r\times\vec p$ is constant; $\vec r$ always stays perpendicular to the fixed vector $\vec\ell\neq0$, so the motion lies in a plane. Correct.

(C) The eccentricity is $e=\sqrt{1+\dfrac{2E\ell^2}{mk^2}}$ (with $k=GMm$). For $E=0$, $e=1$: a parabola. Correct.

(D) Kepler's second law: $\dfrac{dA}{dt}=\dfrac{\ell}{2m}$ is constant, so \[ T=\frac{S}{\ell/2m}=\frac{2mS}{\ell}. \] Correct. \[ \boxed{(B),\ (C),\ (D)} \]

Answer: (B), (C), (D)

CM04.20IIT JAM 2016, Q4MCQ | 1 mark

Consider a particle of mass $m$ following a trajectory given by $x=x_0\cos\omega_1 t$ and $y=y_0\sin\omega_2 t$, where $x_0, y_0, \omega_1$ and $\omega_2$ are constants of appropriate dimensions. The force on the particle is

central only if $\omega_1=\omega_2$
central only if $x_0=y_0$ and $\omega_1=\omega_2$
always central
central only if $x_0=y_0$ and $\omega_1\neq\omega_2$

Show solution

Differentiating twice, \[ F_x=m\ddot x=-m\omega_1^2x,\qquad F_y=m\ddot y=-m\omega_2^2 y . \] The force is central (directed along $\vec r$ towards the origin) only if $\vec F\parallel\vec r$ at all times, i.e. $F_x/F_y=x/y$, which requires $\omega_1^2=\omega_2^2$. Then $\vec F=-m\omega^2\vec r$ for any $x_0,y_0$ (an elliptical orbit is allowed). \[ \boxed{\text{central only if }\omega_1=\omega_2} \]

Answer: (A) central only if $\omega_1=\omega_2$

CM04.21IIT JAM 2016, Q59NAT | 2 marks

The maximum and minimum speeds of a comet that orbits the Sun are $80$ and $10\,km/s$ respectively. The ratio of the aphelion distance of the comet to the radius of the Earth's orbit is .

(Assume that Earth moves in a circular orbit of radius $1.5\times10^{8}\,km$ with a speed of $30\,km/s$)

Show solution

For the Earth's circular orbit, $GM=v_E^2R_E$.

For the comet, angular momentum conservation between perihelion ($v_p=80$) and aphelion ($v_a=10$) gives \[ v_pr_p=v_ar_a\;\Rightarrow\;r_p=\frac{r_a}{8}. \] Energy conservation: \[ \frac12v_p^2-\frac{GM}{r_p}=\frac12v_a^2-\frac{GM}{r_a} \;\Rightarrow\; \frac12\left(v_p^2-v_a^2\right)=GM\left(\frac{8}{r_a}-\frac{1}{r_a}\right)=\frac{7v_E^2R_E}{r_a}. \] \[ \frac{r_a}{R_E}=\frac{14\,v_E^2}{v_p^2-v_a^2}=\frac{14\times900}{6400-100}=\frac{12600}{6300}=\boxed{2} \]

Answer: 2 (range 1.9 to 2.1)

CM04.22GATE 2026, Q56MSQ | 2 marks

An $\alpha$ particle moves towards a fixed nucleus carrying charge $Ze$, with initial speed $v_0$ and impact parameter $b$. Starting from a large distance from the nucleus, its distance of closest approach is $r_m$ and its speed there is $v_m$. Then which of the following options is/are correct? $\left(k=\dfrac1{4\pi\epsilon_0}\text{ and } r_0=k\dfrac{Ze^2}{mv_0^2}\right)$

$v_0b=v_mr_m$
$v_0b=2v_mr_0$
For $\dfrac b{r_0}\ll1$, $r_m=4r_0+\dfrac{b^2}{2r_0}$ ignoring higher order corrections in $\dfrac b{r_0}$
For $\dfrac b{r_0}\ll1$, $r_m=4r_0+\dfrac{b^2}{8r_0}$ ignoring higher order corrections in $\dfrac b{r_0}$

Show solution

Angular momentum. The force is central, so $mv_0b=mv_mr_m$, and at closest approach the velocity is perpendicular to $\vec r$. Hence (A) is true.

Energy. The $\alpha$ particle has charge $2e$, so $U(r)=\dfrac{2kZe^2}{r}=\dfrac{2mv_0^2r_0}{r}$. Then \[ \tfrac12mv_0^2=\tfrac12mv_m^2+\frac{2mv_0^2r_0}{r_m},\qquad v_m=\frac{v_0b}{r_m} \;\Rightarrow\;r_m^2-4r_0r_m-b^2=0, \] \[ r_m=2r_0+\sqrt{4r_0^2+b^2}\approx4r_0+\frac{b^2}{4r_0}\qquad(b\ll r_0). \] The correction term is $b^2/4r_0$, which matches neither (C) ($b^2/2r_0$) nor (D) ($b^2/8r_0$). Option (B) would require $r_m=2r_0$, which is below even the head-on value $4r_0$. So only (A) is correct. \[\boxed{\text{(A) only}}\]

Answer: (A)

CM04.23GATE 2022, Q49MSQ | 2 marks

A particle of unit mass moves in a potential $V(r)=-V_0e^{-r^2}$. If the angular momentum of the particle is $L=0.5\sqrt{V_0}$, then which of the following statements are true?

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Chapter 4 Central Forces and Gravitation Previous Year Questions with Solutions

CM04: Central Forces and Gravitation

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