Kepler’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)
1. $GMK = 4\pi^2$
2. $K = G$
3. $K = \frac{1}{G}$
4. $GM = 4\pi^2$
View Answer
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 $GMK = 4\pi^2$.
A geostationary satellite is orbiting the earth at a height of $5R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is:
(2012 Pre)
1. $5$
2. $10$
3. $6\sqrt{2}$
4. $\sqrt{2}$
View Answer
For the geostationary satellite, $T_1 = 24\text{ hours}$, $r_1 = R + 5R = 6R$. For the second satellite, $r_2 = R + 2R = 3R$. Using Kepler's third law $T^2 \propto r^3$, we have $T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{6R}\right)^{3/2} = 24 \left(\frac{1}{2}\right)^{3/2} = 6\sqrt{2}\text{ hours}$.
A 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)
1. $\left(\frac{r_1}{r_2}\right)^2$
2. $\frac{r_2}{r_1}$
3. $\left(\frac{r_2}{r_1}\right)^2$
4. $\frac{r_1}{r_2}$
View Answer
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 $\frac{r_2}{r_1}$.
The largest and the shortest distance of the earth from the sun are $r_1$ and $r_2$. Its distance from the sun when it is at perpendicular to the major axis of the orbit drawn from the sun is:
(1988)
1. $\frac{r_1+r_2}{4}$
2. $\frac{r_1+r_2}{r_1-r_2}$
3. $\frac{2r_1r_2}{r_1+r_2}$
4. $\frac{r_1+r_2}{3}$
View Answer
The distance from the sun when the planet is perpendicular to the major axis drawn from the sun is the semi-latus rectum of the elliptical orbit. It is calculated as the harmonic mean of the apoapsis and periapsis distances, giving $\frac{2r_1r_2}{r_1+r_2}$.
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
(2018)
1. Time period of a simple pendulum on the Earth would decrease
2. Walking on the ground would become more difficult
3. Raindrops will fall faster
4. '$g$' on the Earth will not change
View Answer
The value of acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$. If $G$ increases by $10$ times, $g$ also increases by $10$ times since it depends on the mass of the Earth, not the Sun. Hence, the statement that '$g$' will not change is incorrect.
Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is:
(1995)
1. $\frac{1}{2}\sqrt{\frac{Gm}{R}}$
2. $\sqrt{\frac{4Gm}{R}}$
3. $\frac{1}{2R}\sqrt{\frac{1}{Gm}}$
4. $\sqrt{\frac{Gm}{R}}$
View Answer
The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.
The earth (mass $= 6 \times 10^{24}\text{ kg}$) revolves around the sun with an angular velocity of $2 \times 10^{-7}\text{ rad/s}$ in a circular orbit of radius $1.5 \times 10^8\text{ km}$. The force exerted by the sun on the earth, in newton, is:
(1995)
1. $36 \times 10^{21}$
2. $27 \times 10^{39}$
3. Zero
4. $18 \times 10^{25}$
View Answer
The force is the centripetal force $F = mR\omega^2$. Substituting the values: $F = (6 \times 10^{24}) \times (1.5 \times 10^{11}\text{ m}) \times (2 \times 10^{-7})^2 = 9 \times 10^{35} \times 4 \times 10^{-14} = 36 \times 10^{21}\text{ N}$.