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 equal masses \(m\) and \(m\) are hung from a balance whose scale pans differ in vertical height by \(h\). The error in weighing in terms of density of the earth \(\rho\) is :
The change in gravity over height \(h\) is \(dg = \frac{2g}{R} h\). Using \(g = \frac{4}{3} \pi G \rho R\), we get \(dg = \frac{8}{3} \pi G \rho h\). Thus, the difference in weight (error) is \(m \cdot dg = \frac{8}{3} \pi G \rho m h\).
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?
At the point where the field is zero, \(\frac{Gm}{r_1^2} = \frac{GM}{r_2^2}\), which gives \(r_1 = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d\) and \(r_2 = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d\). The potential is \(V = -\frac{Gm}{r_1} - \frac{GM}{r_2} = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\).
Gravitational potential difference between a point on surface of planet and another point \(10\text{ m}\) above is \(4\text{ J/kg}\). Considering gravitational field to be uniform, how much work is done in moving a mass of \(2.0\text{ kg}\) from the surface to a point \(5.0\text{ m}\) above the surface?
For a uniform field, potential difference is proportional to distance. Thus, \(\Delta V' = \frac{5}{10} \times 4 = 2\text{ J/kg}\). The work done is \(W = m \Delta V' = 2.0 \times 2 = 4.0\text{ J}\).
Two concentric shells have mass \(M\) and \(m\) and their radii are \(R\) and \(r\) respectively, where \(R > r\). What is the gravitational potential at their common centre ?
The gravitational potential inside any spherical shell is constant and equals the potential at its surface. Therefore, the total potential at the common centre is the sum of the potentials: \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).
The gravitational force between two particles with masses \(m\) and \(M\), initially at rest at great separation, pulls them together. When their separation becomes \(d\), then speed of either particle relative to the other will be :
By conservation of mechanical energy, the relative speed is found using the reduced mass \(\mu = \frac{mM}{m+M}\). Thus, \(\frac{1}{2} mu v_{\text{rel}}^2 = \frac{GMm}{d}\), which simplifies to \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).
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 \(v'_e = \sqrt{\frac{2GM}{R/4}} = 2v_e = 2 \times 11.2 = 22.4\text{ km/s}\).
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 \(\Delta U = U_f - U_i = -\frac{GMm}{R + nR} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} \left(1 - \frac{1}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR\).
The escape velocity for a planet is \(v_e\). A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
Using conservation of mechanical energy from infinity to the centre: \(0 = \frac{1}{2}mv^2 - \frac{3GmM}{2R}\). Since escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\), we get \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).
The mass of a spaceship is \(1000\text{ kg}\). It is to be launched from the earth’s surface out into free space. The value of \(‘g’\) and \(‘R’\) (radius of earth) are \(10\text{ m/s}^2\) and \(6400\text{ km}\) respectively. The required energy for this work will be :
The minimum energy required to escape the earth's gravitational pull from the surface is \(E = \frac{GMm}{R} = mgR\). Substituting \(m = 1000\text{ kg}\), \(g = 10\text{ m/s}^2\), and \(R = 6.4 \times 10^6\text{ m}\), we find \(E = 6.4 \times 10^{10}\text{ Joules}\).