If the earth be at one half its present distance from the sun, number of days in the year will be nearly
According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).
The period of a satellite in a circular orbit of radius \(R\) is \(T\). What is the period of another satellite in a circular orbit of radius \(4R\) ?
By Kepler's Third Law, \(T^2 \propto R^3\). Therefore, \(\frac{T'}{T} = \left(\frac{4R}{R}\right)^{3/2} = 8\), which gives \(T' = 8T\).
Two satellites S and S’ revolve around the earth at distances \(3R\) and \(6R\) from the centre of earth. Their periods of revolution will be in the ratio
Using Kepler's Third Law, \(T^2 \propto r^3 \Rightarrow \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{3R}{6R}\right)^{3/2} = \left(\frac{1}{2}\right)^{1.5} = \frac{1}{2^{1.5}}\). Hence, the ratio is 1 : \(2^{1.5}\).
A satellite revolves around a planet in an elliptical orbit of minor and major axes \(a\) and \(b\) respectively. If T be the time period of the satellite, then \(T^2\) is proportional to
According to Kepler's Third Law, \(T^2\) is proportional to the cube of the semi-major axis. Since the major axis is given as \(b\), the semi-major axis is \(b/2\), making \(T^2 \propto b^3\).
A geostationary satellite has an orbital period of
A geostationary satellite remains stationary relative to the Earth's surface, meaning its orbital period must equal the rotation period of the Earth, which is 24 hours.
Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \(r^{-5/2}\), then the square of the time period will be proportional to
The centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).
Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is :
Areal velocity is given by \(\frac{dA}{dt} = \frac{L}{2m} = \frac{vr}{2} = \frac{\sqrt{GMr}}{2}\). Since areal velocity is proportional to \(\sqrt{r}\), the ratio is \(\sqrt{\frac{R}{4R}} = \frac{1}{2}\) or 1 : 2.
Two bodies, each of mass \(M\), are kept fixed with a separation \(2L\). A particle of mass \(m\) is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is \(G\). The correct statement(s) is (are) :
(a) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(4\sqrt{\frac{GM}{L}}\)
(b) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(2\sqrt{\frac{GM}{L}}\)
(c) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(\sqrt{\frac{2GM}{L}}\)
(d) The energy of the mass \(m\) remains constant
At the midpoint, potential energy is \(U = -\frac{2GMm}{L}\). For escaping to infinity, total mechanical energy must be at least 0:
\(\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 ⇒ v = 2\sqrt{\frac{GM}{L}}\). Mechanical energy remains conserved as only gravity acts.
The gravitational potential energy of a body of mass \(m\) at the earth’s surface is \(-mgR_e\). Its gravitational potential energy at a height \(R_e\) from the earth’s surface will be (Here \(R_e\) is the radius of the earth)
At the surface, \(U_s = -\frac{GMm}{R_e} = -mgR_e\). At a height \(h = R_e\), the distance from the center is \(r = 2R_e\). Thus, \(U_h = -\frac{GMm}{2R_e} = \frac{-mgR_e}{2}\).
Two identical satellites are at height \(R\) and \(7R\) from earth surface, the ratio of their kinetic energies will be :
Kinetic energy of a satellite is \(K = \frac{GMm}{2r}\). Here \(r_1 = R + R = 2R\) and \(r_2 = R + 7R = 8R\). The ratio \(frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4\).