If the earth stops rotating about its axis, the acceleration due to gravity will remain unchanged at
The effective acceleration due to gravity at latitude \(\lambda\) is \(g' = g - \omega^2 R \cos^2 \lambda\). At the poles, \(\lambda = 90^\circ\), so \(g' = g\). Thus, rotation has no effect at the poles.
One goes from the centre of the earth to an altitude half the radius of the earth, where will the \(g\) be greatest ?
Inside the earth, \(g\) increases linearly from center to surface: \(g(r) \propto r\). Outside, it decreases: \(g(r) \propto 1/r^2\). Thus, \(g\) is maximum at the surface of the earth.
An iron sphere and an aluminium sphere, both of same radius are dropped from the top of a tower 100m high. At a height 40 m above the ground, both of them will have same.
In free fall under gravity (neglecting air resistance), all bodies experience the same acceleration due to gravity \(g\), regardless of their mass or density.
Two bodies of masses \(m\) and \(M\) are placed at distance \(d\) apart. What is the gravitational potential (\(V\)) at the position where the gravitational field due to them is zero is \(V\) :
Let the point of zero field be at distance \(r_1\) from \(m\) and \(r_2\) from \(M\). Then \(\frac{\sqrt{m}}{r_1} = \frac{\sqrt{M}}{r_2}\), with \(r_1 + r_2 = d\). Solving gives \(r_1 = \frac{\sqrt{m}d}{\sqrt{m}+\sqrt{M}}\) and \(r_2 = \frac{\sqrt{M}d}{\sqrt{m}+\sqrt{M}}\). Thus, \(V = -\frac{Gm}{r_1} - \frac{GM}{r_2} = -\frac{G}{d}(\sqrt{m}+\sqrt{M})^2\).
The gravitational force of attraction between two bodies is \(F\) newtons. If the mass of each body and the distance between them are doubled, then the gravitational force between them in newton is
Formula of gravitational force is \(F = \frac{G m_1 m_2}{r^2}\). If masses and distance are doubled: \(F' = \frac{G(2m_1)(2m_2)}{(2r)^2} = \frac{4 G m_1 m_2}{4 r^2} = F\). Thus, the force remains unchanged.
If escape velocity from earth is \(11.2\text{ km/s}\), Then escape velocity from a planet of mass as that of earth but of its one fourth radius
Escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\). Since the mass of the planet is equal to that of Earth but the radius is \(R/4\), the escape velocity will be \(v'_e = \sqrt{\frac{2GM}{R/4}} = 2v_e = 2 \times 11.2 = 22.4\text{ km/s}\).
A tunnel is dug along the diameter of the earth (radius \(R\) and mass \(M\)). There is a particle of mass \(‘m’\) at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches to the surface of the earth is:
By conservation of mechanical energy, \(K_{\text{centre}} + U_{\text{centre}} = K_{\text{surface}} + U_{\text{surface}}\). With \(K_{\text{surface}} = 0\), we get \(\frac{1}{2}mv^2 - \frac{3GmM}{2R} = -\frac{GmM}{R}\), which gives \(v = \sqrt{\frac{GM}{2R}}\).
What is the increase in gravitational potential energy of an object of mass \(m\) raised from the surface of earth to a height equal to \(n\) times of earth radius ?
The increase in potential energy is \(\Delta U = U_f - U_i = -\frac{GMm}{R + nR} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} \left(1 - \frac{1}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR\).
One goes from the centre of the earth to an altitude half the radius of the earth, where will the \(g\) be greatest ?
The acceleration due to gravity is zero at the centre, increases linearly inside the earth up to the surface where it is maximum \(g = \frac{GM}{R^2}\), and then decreases as \(1/r^2\) outside the surface. Thus, \(g\) is greatest at the surface.
An iron sphere and an aluminium sphere, both of same radius are dropped from the top of a tower \(100\text{ m}\) high. At a height \(40\text{ m}\) above the ground, both of them will have same:
In the absence of air resistance, all falling bodies have the same acceleration due to gravity \(g\) independent of their mass. Since their masses are different due to different densities, other quantities will differ.