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. $GM = 4\pi^2$
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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 $GMK = 4\pi^2$.
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. $\frac{r_1+r_2}{3}$
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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 $\frac{2r_1r_2}{r_1+r_2}$.
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. $\sqrt{2}$
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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. $\frac{r_1}{r_2}$
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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 $\frac{r_2}{r_1}$.
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
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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.
A spherical planet has a mass $M_P$ and diameter $D_P$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to:
(2012 Pre)
1. $\frac{4GM_P}{D_P^2}$
2. $\frac{GM_Pm}{D_P^2}$
3. $\frac{GM_P}{D_P^2}$
4. $\frac{4GM_Pm}{D_P^2}$
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Acceleration due to gravity is given by $g = \frac{GM_P}{R_P^2}$. Substituting the radius as half of the diameter, $R_P = \frac{D_P}{2}$, we get $g = \frac{GM_P}{(D_P/2)^2} = \frac{4GM_P}{D_P^2}$.
Imagine a new planet having the same density as that of earth but it is $3$ times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the new planet is $g’$, then:
(2005)
1. $g' = 3g$
2. $g' = \frac{g}{9}$
3. $g' = 9g$
4. $g' = \frac{g}{3}$
View Answer
Acceleration due to gravity in terms of density is $g = \frac{4}{3}\pi \rho G R$. Since density $\rho$ is constant, $g \propto R$. For a planet $3$ times bigger in size ($R' = 3R$), the new gravity is $g' = 3g$.