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 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 \(A = \frac{L}{2m} T ⇒ L = \frac{2mA}{T}\).
For energy of satellite, match the columns (symbols have their respective meaning):
**Column-I**
(i) Kinetic energy
(ii) Potential energy
(iii) Total energy
**Column-II**
(p) \(\frac{L^2}{2mr^2}\)
(q) \(-\frac{L^2}{mr^2}\)
(r) \(-\frac{L^2}{2mr^2}\)
Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{L^2}{2mr^2}\) (since \(L = mvr\)). Potential energy \(U = -\frac{GMm}{r} = -\frac{L^2}{mr^2}\). Total energy \(E = K + U = -\frac{L^2}{2mr^2}\). Thus (i)-p, (ii)-q, (iii)-r.
Assertion : If rotation of earth about its own axis is suddenly stopped then acceleration due to gravity will increase at all places on the earth (except poles).
Reason : At height \(h\) from the surface of earth, acceleration due to gravity is \(g_h = g \left(1 – \frac{2h}{R_e}\right)\) (If \(h \ll R_e\)) [\(R_e \rightarrow\) radius of earth]
The effective gravity is \(g' = g - \omega^2 R \cos^2\lambda\). If rotation stops (\(\omega = 0\)), \(g'\) increases everywhere except the poles (where \(\lambda = 90^\circ\)). Thus Assertion is true. The Reason is also a true independent formula for gravity at a height, but not the explanation.
The gravitational field in a region is given by \(vec{I} = 10(\hat{i} + \hat{j})\text{ N/kg}\). The work done by gravitational field to shift a particle of mass \(2\text{ kg}\) from position \((0,0)\) to \((5, 4)\) will be :
The force is \(\vec{F} = mvec{I} = 2 \times 10(\hat{i} + \hat{j}) = 20\hat{i} + 20\hat{j}\text{ N}\). Displacement is \(\vec{d} = 5\hat{i} + 4\hat{j}\text{ m}\). Work done \(W = \vec{F} \cdot \vec{d} = (20)(5) + (20)(4) = 180\text{ J}\).
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 \(frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4\).
The weight of a body at a distance \(r\) (\(r > R\)) from the centre of earth is \(W\). The weight at a distance \(2r\) from the centre of earth is : (\(R \rightarrow\) Radius of earth)
Outside the earth, weight \(W \propto \frac{1}{r^2}\). When distance is doubled to \(2r\), the weight becomes \(\frac{1}{2^2} = \frac{1}{4}\) of its initial value, i.e., \(\frac{W}{4}\).