Q1. In a two-body system with masses \(m_1\) and \(m_2\), the relative position vector is usually defined as:

\(\vec r=\vec r_1+\vec r_2\)
\(\vec r=\vec r_1-\vec r_2\) (or equivalently the opposite convention)
\(\vec r=m_1\vec r_1+m_2\vec r_2\)
\(\vec r=\vec r_1\cdot\vec r_2\)

Q2. The centre-of-mass position vector of two particles is:

\(\displaystyle \frac{\vec r_1+\vec r_2}{2}\)
\(\displaystyle \vec R=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}\)
\(\displaystyle \vec R=m_1\vec r_1+m_2\vec r_2\)
\(\displaystyle \vec R=\vec r_1-\vec r_2\)

Q3. The reduced mass of a two-body system is:

\(m_1+m_2\)
\(m_1-m_2\)
\(\displaystyle \mu=\frac{m_1m_2}{m_1+m_2}\)
\(\displaystyle \mu=\frac{m_1+m_2}{m_1m_2}\)

Q4. Two particles have masses \(2\,\text{kg}\) and \(3\,\text{kg}\). Their reduced mass is:

\(5\,\text{kg}\)
\(2.5\,\text{kg}\)
\(1.2\,\text{kg}\)
\(6\,\text{kg}\)

Q5. If \(m_1=m_2=m\), the reduced mass is:

\(m\)
\(2m\)
\(m/2\)
\(m/4\)

Q6. If \(m_1 \gg m_2\), the reduced mass is approximately:

\(m_1\)
\(m_2\)
\(m_1+m_2\)
Zero

Q7. The two-body central force problem can be reduced to:

Two independent three-body problems
Centre-of-mass motion and an equivalent one-body problem of reduced mass \(\mu\)
A system with zero mass
A harmonic oscillator in every case

Q8. In the absence of an external force, the centre of mass of an isolated two-body system:

Must remain at rest
Moves in a circle
Moves with constant velocity
Accelerates toward the heavier body

Q9. A central force is a force which:

Is always constant
Acts along the line joining the particle and the force centre
Is always perpendicular to the position vector
Acts only in circular motion

Q10. A central force can be written in the form:

\(\vec F=F(r)\hat{\theta}\)
\(\vec F=F(r)\hat r\)
\(\vec F=F(t)\hat{\theta}\)
\(\vec F=F(r)\hat z\)

Q11. The torque about the force centre due to a central force is:

Maximum
\(rF\)
Zero
\(rF/2\)

Q12. Since the torque due to a central force is zero, which quantity is conserved?

Linear momentum
Radial velocity
Angular momentum
Acceleration

Q13. Conservation of angular momentum in central force motion implies that the motion:

Must be along a straight line
Takes place in a plane
Must be circular
Must be three-dimensional

Q14. The angular momentum of the equivalent particle in plane polar coordinates is:

\(L=\mu r\dot r\)
\(L=\mu r^2\dot\theta\)
\(L=\mu\dot\theta/r^2\)
\(L=\mu r^2\ddot\theta\)

Q15. The areal velocity of a particle moving under a central force is:

\(r\dot\theta\)
\(\displaystyle \frac{dA}{dt}=\frac12r^2\dot\theta\)
\(r^2\ddot\theta\)
\(\dot r\dot\theta\)

Q16. Using \(L=\mu r^2\dot\theta\), the areal velocity is:

\(L/\mu\)
\(2L/\mu\)
\(L/(2\mu)\)
\(\mu/(2L)\)

Q17. The constancy of areal velocity under a central force is the basis of:

Kepler's first law
Kepler's second law
Kepler's third law
Newton's third law

Q18. The radial equation of motion under a central force is:

\(\mu(\ddot r+r\dot\theta^2)=F(r)\)
\(\mu(\ddot r-r\dot\theta^2)=F(r)\)
\(\mu r\ddot\theta=F(r)\)
\(\mu\ddot r=0\)

Q19. The transverse equation for central force motion is:

\(r\ddot\theta-\dot r\dot\theta=0\)
\(r\ddot\theta+2\dot r\dot\theta=0\)
\(\ddot\theta+r\dot\theta=0\)
\(r\ddot\theta=F(r)\)

Q20. The effective potential for central force motion is:

\(V_{\text{eff}}=V(r)-\frac{L^2}{2\mu r^2}\)
\(V_{\text{eff}}=V(r)+\frac{L^2}{2\mu r^2}\)
\(V_{\text{eff}}=V(r)+Lr\)
\(V_{\text{eff}}=V(r)+\frac12\mu r^2\)

Q21. The term \(\frac{L^2}{2\mu r^2}\) in the effective potential is called:

Gravitational potential
Elastic potential
Centrifugal potential term
Rest energy

Q22. The total energy of the equivalent one-body central force problem can be written as:

\(E=\frac12\mu\dot r^2-V_{\text{eff}}\)
\(E=\frac12\mu\dot r^2+V_{\text{eff}}(r)\)
\(E=\mu\dot r+V\)
\(E=Lr+V\)

Q23. In deriving the differential equation of an orbit, it is convenient to introduce:

\(u=r^2\)
\(u=1/r\)
\(u=\dot r\)
\(u=r\theta\)

Q24. Binet's orbit equation for a central force is:

\(\displaystyle \frac{d^2u}{d\theta^2}-u=-\frac{\mu}{L^2u^2}F(1/u)\)
\(\displaystyle \frac{d^2u}{d\theta^2}+u=-\frac{\mu}{L^2u^2}F(1/u)\)
\(\displaystyle \frac{du}{d\theta}+u=F(r)\)
\(\displaystyle \frac{d^2u}{dt^2}+u=0\)

Q25. For an attractive inverse-square force, \(F(r)=-k/r^2\), Binet's equation becomes:

\(\displaystyle u''+u=-\frac{\mu k}{L^2}\)
\(\displaystyle u''+u=\frac{\mu k}{L^2}\)
\(u''-u=0\)
\(u''+u=kr^2\)

Q26. The orbit under an attractive inverse-square force can be written as:

\(r=a\theta\)
\(\displaystyle r=\frac{p}{1+e\cos\theta}\)
\(r=p(1+e\theta)\)
\(r=pe^\theta\)

Q27. In the orbit equation \(r=p/(1+e\cos\theta)\), \(e\) represents:

Energy
Angular momentum
Eccentricity
Areal velocity

Q28. An inverse-square orbit with \(e=0\) is:

A circle
An ellipse
A parabola
A hyperbola

Q29. An orbit with \(0<e<1\) is:

Circle only
Ellipse
Parabola
Hyperbola

Q30. An orbit with \(e=1\) is:

Circle
Ellipse
Parabola
Hyperbola

Q31. An orbit with \(e>1\) is:

Circle
Ellipse
Parabola
Hyperbola

Q32. For the attractive inverse-square potential with \(V(\infty)=0\), a negative total energy corresponds to:

A bound orbit
A parabolic escape orbit
A hyperbolic orbit
No possible motion

Q33. For an inverse-square attractive force, zero total energy corresponds to:

Circle
Ellipse
Parabola
Hyperbola

Q34. For an inverse-square attractive force, positive total energy corresponds to:

Circle
Ellipse
Parabola
Hyperbola

Q35. A circular orbit of radius \(r_0\) corresponds to:

A maximum of \(V_{\text{eff}}\) in every case
A stationary point of the effective potential
\(L=0\) necessarily
Infinite kinetic energy

Q36. The condition for a circular orbit at \(r=r_0\) is:

\(V_{\text{eff}}(r_0)=0\)
\(\displaystyle \left.\frac{dV_{\text{eff}}}{dr}\right|_{r_0}=0\)
\(\displaystyle \left.\frac{dV_{\text{eff}}}{dr}\right|_{r_0}=1\)
\(V(r_0)=\infty\)

Q37. A circular orbit is stable if:

\(\displaystyle \frac{d^2V_{\text{eff}}}{dr^2}<0\)
\(\displaystyle \frac{d^2V_{\text{eff}}}{dr^2}>0\)
\(V_{\text{eff}}=0\) only
\(L=0\)

Q38. A minimum of the effective potential represents:

An unstable circular orbit
A stable circular orbit
A parabolic orbit only
An impossible orbit

Q39. According to Bertrand's theorem, the two central-force potentials for which all bound orbits are closed are proportional to:

\(r\) and \(1/r^2\)
\(r^2\) and \(-1/r\)
\(r^3\) and \(1/r^3\)
\(\ln r\) and \(r\)

Q40. Kepler's first law states that a planet moves:

In a circular orbit with the Sun at its centre
In an elliptical orbit with the Sun at one focus
In a parabolic orbit
With constant linear velocity

Q41. Kepler's second law states that the line joining a planet and the Sun:

Has constant length
Sweeps out equal areas in equal intervals of time
Rotates with constant angular velocity
Always remains perpendicular to the velocity

Q42. According to Kepler's third law:

\(T\propto a\)
\(T^2\propto a^2\)
\(T^2\propto a^3\)
\(T^3\propto a^2\)

Q43. A planet has a semi-major axis four times that of another planet. According to Kepler's third law, the ratio of their orbital periods is:

2
4
8
16

Q44. If the semi-major axis of a planetary orbit becomes \(9\) times larger, its orbital period becomes:

3 times
9 times
27 times
81 times

Q45. For a circular orbit of radius \(r\) around a mass \(M\), equating gravitational and centripetal forces gives:

\(\displaystyle \frac{GMm}{r}=\frac{mv^2}{r}\)
\(\displaystyle \frac{GMm}{r^2}=\frac{mv^2}{r}\)
\(GMm=mv^2r^2\)
\(\displaystyle \frac{GMm}{r^3}=mv^2\)

Q46. The orbital speed of a body in a circular orbit of radius \(r\) around a mass \(M\) is:

\(\sqrt{GMr}\)
\(\sqrt{\frac{GM}{r}}\)
\(GM/r\)
\(\sqrt{GM/r^2}\)

Q47. A satellite moves in a circular orbit of radius \(r\) with speed \(v\). If the orbital radius becomes \(4r\), its circular orbital speed becomes:

\(4v\)
\(2v\)
\(v/2\)
\(v/4\)

Q48. For a circular planetary orbit, Kepler's third law and Newton's gravitation give:

\(\displaystyle T^2=\frac{GM}{4\pi^2}r^3\)
\(\displaystyle T^2=\frac{4\pi^2}{GM}r^3\)
\(\displaystyle T=\frac{4\pi^2r^3}{GM}\)
\(\displaystyle T^2=4\pi^2GMr^3\)

Q49. Suppose a planet moves in a circular orbit and Kepler's third law gives \(T^2\propto r^3\). The centripetal acceleration is therefore proportional to:

\(r\)
\(1/r\)
\(1/r^2\)
\(r^2\)

Q50. A planet moves in a circular orbit of radius \(r\) with period \(T\). Using Kepler's third law, the central force required to maintain the orbit varies with \(r\) as:

\(F\propto r\)
\(F\propto r^2\)
\(F\propto 1/r\)
\(F\propto 1/r^2\)