An 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, \( U = 2E_0 \).
Two 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, \( \frac{g_1}{g_2} = \frac{R_1}{R_2} \times \frac{\rho_1}{\rho_2} = \frac{2}{3} \times \frac{3}{2} = 1 \).
The 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 \(h = 2R\).
A 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 \(t = \sqrt{\frac{2h}{3g}}\).
Assertion: 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.
A 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.
The acceleration due to gravity (on earth) depends upon
Acceleration due to gravity on the surface of earth is given by \(g = \frac{G M_e}{R_e^2}\). Hence, it depends on the mass of the earth, not on the mass or size of the falling body.
An 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 a solid sphere, the acceleration due to gravity at a distance \(r\) from the center is \(g' = \frac{g r}{R}\). For \(r = R/2\), we have \(g' = \frac{g (R/2)}{R} = g/2\).
A thin rod of length \(L\) is bent to form a circle. Its mass is \(M\). What force will act on the mass \(m\) placed at the centre of the circle ?
Due to symmetry, the gravitational forces exerted by all symmetric parts of the ring at the center cancel each other out, resulting in a net force of zero.
A spherical shell has mass \(M\) and radius \(R\). A point mass \(m/2\) kept inside the shell at a distance \(R/2\) from centre. Then force of attraction on the mass is:
The gravitational field inside a uniform spherical shell is zero everywhere. Therefore, the gravitational force on any mass kept inside the shell is zero.