Question 11:
easyA planet of mass \(m\) is moving in an elliptical orbit about the sun with time period \(T\). If \(A\) be the area of orbit, then its angular momentum would be :
According to Kepler's second law, the areal velocity of the planet is constant and is given by \(\frac{dA}{dt} = \frac{L}{2m}\). Integrating over one time period \(T\) gives \(A = \frac{L}{2m} T ⇒ L = \frac{2mA}{T}\).
Question 12:
easyAssertion (A): Period of revolution of satellite in circular orbit around earth is inversely proportional to cube of its orbital speed.
Reason (R): Period of revolution in uniform circular motion is given by \( T = \frac{2\pi r}{v} \) where \( r \) is radius of orbit and \( v \) is speed.
Assertion (A) is true. For a satellite in circular orbit, orbital speed \( v = \sqrt{\frac{GM}{r}} \) implying \( r \propto \frac{1}{v^2} \). The period is \( T = \frac{2\pi r}{v} \). Substituting \( r \), we get \( T \propto \frac{1/v^2}{v} \propto \frac{1}{v^3} \).
Reason (R) is true. The formula \( T = \frac{2\pi r}{v} \) is the correct definition for the period of uniform circular motion. However, (R) is a kinematic definition and does not explain the dynamic relationship between \( T \) and \( v \) for a satellite, which requires considering gravity. Thus, (R) is not the correct explanation of (A).
Question 13:
easyAssertion (A): When planet moves in elliptical orbit around Sun. Its angular momentum about sun remains conserved.
Reason (R): Total mechanical energy of planet – sun system remains conserved.
Assertion (A) is true. Gravitational force is a central force, so the torque about the Sun is zero, leading to angular momentum conservation. Reason (R) is also true. Gravitational force is conservative, so total mechanical energy of the system is conserved. However, energy conservation does not explain angular momentum conservation, as they are distinct conservation laws.
Question 14:
easyAssertion (A): The radius vector from the sun to a planet sweeps out equal areas in equal times interval.
Reason (R): Transverse (perpendicular to radius vector) acceleration of the planet is zero.
Assertion (A): This statement is Kepler's Second Law, which is a direct consequence of angular momentum conservation. So (A) is true.nReason (R): For a central force, like gravity, the force acts along the radius vector, meaning no transverse force component exists. Thus, transverse acceleration is zero. So (R) is true.n(R) explains (A) because zero transverse acceleration leads to conservation of angular momentum, which implies Kepler's Second Law.
Question 15:
easyGiven below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Angular momentum conservation can be used to explain Kepler’s second law of planetary motion.
Reason (R): Areal velocity of a planet revolving around the sun is equal to its angular momentum.
In the light of the above statements, choose the correct answer from the options given below:
Assertion (A) is true because Kepler's second law represents the conservation of angular momentum. Reason (R) is false because areal velocity is \(\frac{dA}{dt} = \frac{L}{2m}\), which is proportional to angular momentum \(L\) but not equal to it.
Question 16:
moderateKepler’s third law states that square of period of revolution ($T$) of a planet around the sun, is proportional to third power of average distance $r$ between sun and planet, i.e., $T^2 = Kr^3$ here $K$ is constant. If the masses of sun and planet are $M$ and $m$ respectively then as per Newton’s law of gravitation force of attraction between them is $F = \frac{GMm}{r^2}$ here $G$ is gravitational constant. The relation between $G$ and $K$ is described as:
(2015)
We know that the time period of a planet is given by $T^2 = \frac{4\pi^2}{GM}r^3$. Comparing this with $T^2 = Kr^3$, we get $K = \frac{4\pi^2}{GM}$. Rearranging this gives $GMK = 4\pi^2$.
Question 17:
moderateA planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively, then the ratio is
(2011 Mains)
By the conservation of angular momentum at the closest and farthest points, $mv_1r_1 = mv_2r_2$. Therefore, the ratio of their linear velocities $\frac{v_1}{v_2}$ is equal to $\frac{r_2}{r_1}$.
Question 18:
moderateThe period of revolution of planet A around the sun is $8$ times that of B. The distance of A from the sun is how many times greater than that of B from the sun?
(1997)
According to Kepler's third law, $T^2 \propto r^3$. Given $T_A = 8T_B$, so $\left(\frac{r_A}{r_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2 = (8)^2 = 64$. Taking the cube root yields $r_A = 4r_B$.
Question 19:
moderateThe distance of two planets from the sun are $10^{13}\text{ m}$ and $10^{12}\text{ m}$ respectively. The ratio of time periods of the planets is:
(1994, 88)
Using Kepler's 3rd Law $T^2 \propto r^3$, the ratio of time periods is $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2}$. Substituting the distances, we get $\frac{T_1}{T_2} = \left(\frac{10^{13}}{10^{12}}\right)^{3/2} = (10)^{3/2} = 10\sqrt{10}$.
Question 20:
moderateA satellite A of mass $m$ is at a distance of $r$ from the surface of the earth. Another satellite B of mass $2m$ is at a distance of $2r$ from the earth’s centre. Their time periods are in the ratio of:
(1993)
Time period of a satellite is independent of its mass. Assuming the distance $r$ for satellite A was intended to be from the centre (a standard typo in this question format), $r_A = r$ and $r_B = 2r$. Their time period ratio $\frac{T_A}{T_B} = \left(\frac{r_A}{r_B}\right)^{3/2} = \left(\frac{1}{2}\right)^{3/2} = 1:2\sqrt{2}$.