Tag Newton’s laws of motion

Kepler’s law of period

Kepler's law of period

Kepler's law of period for planetary motion

According to Kepler’s laws  of period for planetary motion, the square of the period of revolution of as planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

Mathematically, it can be expressed as T^2 = k a^3 , where is the orbital period of the planet, is the semi-major axis of its orbit, and is a constant that is the same for all planets orbiting the Sun. 

Example

Q.A geostationary satellite is orbiting the earth at a height of 6R above the surface of earth where R is the radius of earth. The time period of another satellite at a distance of 3.5 R from the centre of earth is

(A) 16.96 hours

(B) 8.48 hours

(C) 9.25 hours

(D) 15.76 hours

Solution

We know that , orbital speed of satellite is
v_o=\;\sqrt{\frac{GM}r}\;
where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth
T\;=\frac{dis\tan ce\;}{speed}=\;\frac{2\pi\;r}{v_o}
T\;=\;\frac{2\pi\;r}{\sqrt{\displaystyle\frac{GM}r}}
T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}
T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}\\T^2=\;4\;\pi^2\;\frac{r^3}{GM}
\Rightarrow T^2\;\;\propto\;r^3
This is known as Kepler's Law of period for planetary motion
r= R+ h where R is radius of planet/earth and h is height of satellite from surface of earth
\Rightarrow r_1 = R+ 6R=7R \;and \;r_2= 3.5R
\Rightarrow T\;\propto\;r^\frac32
\frac{T_2}{T_1}=\;\left(\frac{r_2}{r_1}\right)^\frac32
\frac{T_2}{T_1}=\;\left(\frac{3.5R}{7R}\right)^\frac32
\frac{T_2}{T_1}=\;\left(\frac{1}{2}\right)^\frac32
\frac{T_2}{T_1}=\frac1{2\sqrt2}
T_2 =T_1 * \frac1{2\sqrt2}
T_2 =24 * \frac1{2\sqrt2}
T_2 =12* \frac1{\sqrt2}
T_2 =12* 0.707= 8.48 \;hours
(B) is correct option

Conclusion

Kepler’s third law is also known as law of period which helps us to find the time period of orbital motion for various distances from sun. It includes following  applications. 

  1. Predicting Planetary Motion: Kepler’s law of periods allows astronomers to predict the orbital periods of planets based on their distances from the sun. This law has been crucial in the study of our solar system and in the discovery and characterization of exoplanets in other solar systems.

  2. Comparing Orbits: By comparing the orbital periods and semi-major axes of different planets or moons, astronomers can gain insights into the structure and dynamics of planetary systems. For example, comparing the orbital periods of moons around a planet can provide information about their relative distances from the planet.

  3. Verification of Kepler’s Laws: Kepler’s laws of planetary motion played a significant role in the development of Newton’s law of universal gravitation and the laws of motion. They served as a crucial test for Newtonian physics and provided evidence for the gravitational force between celestial bodies.

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ>ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

(A) g\left(\frac\sigma\rho-1\right)\;downwards

(B) g\left(\frac\sigma\rho-1\right)\;upwards

(C) g\left(\frac\sigma\rho\right)\;downwards

(D) g\left(\frac\sigma\rho\right)\;upwards

Solution

Let V = Volume of body

\rho = density of body

\sigma = density of lake water

Thus, m=\;\rho\;V

As the density of liquid is more than that of body( σ > ρ), Therefore the body has tendency to float on the liquid because of larger upwards force in comparison to that of weight of body. Hence, net force will be upwards.

W know that F_{up}=V\;\sigma g\;

According to Newton's 2nd law of motion, we can write

F_{net}= F_{up}-mg (upwards)

F_{net}= V\sigma g -V \rho g

ma = V\sigma g -V \rho g

V\rho a = V\sigma g -V \rho g

a=\frac{V(\sigma-\rho)g}{V\rho}

a=\frac{(\sigma-\rho)g}{\rho}

a=g\left(\frac\sigma\rho-1\right)\;upwards

(B) is correct option

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)