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., \(\frac{W}{4}\).
If mass of a planet is \(M\) and radius is \(x\), then the work to be done to slowly take a mass \(m\) from surface of planet to a height \(4x\) is :
Initial position \(r_i = x\), final position \(r_f = x + 4x = 5x\). Work done \(W = U_f - U_i = -\frac{GMm}{5x} - \left(-\frac{GMm}{x}\right) = \frac{4GMm}{5x}\).
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 \(W = \vec{F} \cdot \vec{d} = (20)(5) + (20)(4) = 180\text{ J}\).
The energy required to put a satellite of mass \(m\) from earth surface into a orbit of radius \(2R\) is \(E_1\). The energy further needed to change the orbit of this satellite from its present orbit to radius \(4R\) is \(E_2\). The ratio \(\frac{E_1}{E_2}\) is (where \(R\) is radius of earth:
The energy required to put a satellite in orbit from earth's surface is \(E_1 = -\frac{GMm}{2(2R)} - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R}\). The energy to change orbit from \(2R\) to \(4R\) is \(E_2 = -\frac{GMm}{2(4R)} - \left(-\frac{GMm}{2(2R)}\right) = \frac{GMm}{8R}\). Thus, \(\frac{E_1}{E_2} = 6\), which gives the ratio \(6 : 1\).
A body weighs \(900\text{ N}\) on the surface of earth. How much will it weigh at a height double the radius of earth?
Using the formula for acceleration due to gravity at height \(h\): \(g' = g\left(\frac{R}{R+h}\right)^2\). For \(h = 2R\), \(g' = \frac{g}{9}\) . Hence, weight at this height is \(W' = \frac{W}{9} = \frac{900\text{ N}}{9} = 100\text{ N}\).
The escape velocity from the Earth’s surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is
Escape velocity is given by \(v_e = R\sqrt{\frac{8\pi G\rho}{3}}\). Since the mass density \(rho\) is the same, \(v_e\) is directly proportional to radius \(R\). Therefore, \(v_e' = 4v\).
A particle of mass \(m\) is projected with a velocity \(v = k V_e\) (\(k < 1\)) from the surface of the earth. (\(V_e = \text{escape velocity}\)) The maximum height above the surface reached by the particle is
By conservation of mechanical energy: \(-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h}\). Since \(v = k \sqrt{\frac{2GM}{R}}\), we substitute to get \(-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}\), which yields \(h = \frac{R k^2}{1-k^2}\).