Question 61:
easyAn artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy \( E_0 \). Its potential energy is:
For a satellite in circular orbit, Potential Energy is \( U = -\frac{GMm}{r} \) and Total Energy is \( E_0 = -\frac{GMm}{2r} \). Thus, Potential Energy is twice the Total Energy, \( U = 2E_0 \).
Question 62:
easyTwo planets of radii in the ratio 2 : 3 are made from the material of density in the ratio 3 : 2. Then the ratio of acceleration due to gravity \( g_1 / g_2 \) at the surface of the two planets will be:
Acceleration due to gravity at the surface of a planet is given by \( g = \frac{4}{3} \pi G R \rho \). Therefore, \( \frac{g_1}{g_2} = \frac{R_1}{R_2} \times \frac{\rho_1}{\rho_2} = \frac{2}{3} \times \frac{3}{2} = 1 \).
Question 63:
easyThe height at which the weight of a body becomes 1/9th its weight on the surface of earth (radius of earth is R):
Since acceleration due to gravity varies with height as \(g' = g \frac{R^2}{(R+h)^2}\), setting \(g' = g/9\) gives \(R+h = 3R\), which simplifies to \(h = 2R\).
Question 64:
difficultA body of superdense material with mass twice the mass of earth but size very small compared to the size of earth starts from rest from \(h \ll R\) above the earth’s surface. It reaches earth in time \(t\). Then:
The acceleration of the earth towards the body is \(a_E = 2g\) and the body towards the earth is \(a_B = g\). The relative acceleration is \(a_{rel} = 3g\). Using \(h = \frac{1}{2} a_{rel} t^2\), we get \(t = \sqrt{\frac{2h}{3g}}\).
Question 65:
easyAssertion: If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases.
Reason: The speed of satellite is a constant quantity.
As a satellite moves to a lower orbit, total energy decreases (becomes more negative) due to dissipation, but kinetic energy increases, so speed increases. The speed of a satellite is not universally constant.
Question 66:
easyA clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth but twice the radius:
Since \(g \propto \rho R\), on a planet with twice the radius and same density, \(g' = 2g\). Pendulum time period decreases (\(T_p = 2\pi\sqrt{l/g}\)), so clock P ticks faster. Spring clock is unaffected.
Question 67:
easyIf the earth stops rotating about its axis, the acceleration due to gravity will remain unchanged at
The effective gravity is \(g' = g - \omega^2 R \cos^2 \theta\). At the poles, \(\theta = 90^\circ\), so \(g' = g\), which is independent of the earth's angular velocity.
Question 68:
easyIf three uniform spheres, each having mass \(M\) and radius \(r\), are kept in such a way that each touches the other two, the magnitude of the gravitational force on any sphere due to the other two is
The distance between centers is \(2r\). Force between any two is \(F = \frac{GM^2}{4r^2}\). Since the angle between the forces is \(60^\circ\), the resultant force is \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3} GM^2}{4r^2}\).
Question 69:
easyThe acceleration due to gravity (on earth) depends upon
The acceleration due to gravity is \(g = \frac{GM_e}{R_e^2}\). This depends on the mass of the Earth, not on the properties of the body itself.
Question 70:
easyAn object is placed at a distance of \(R/2\) from the centre of earth. Knowing mass is distributed uniformly, acceleration of that object due to gravity at that point is: (\(g =\) acceleration due to gravity on the surface of earth and \(R\) is the radius of earth)
Inside the earth, the acceleration due to gravity varies linearly with distance: \(g' = \frac{gr}{R}\). For \(r = R/2\), we have \(g' = \frac{g}{2}\).