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 the symmetrical distribution of mass in a circular ring, the gravitational field at the center is zero. Therefore, the force on any mass placed at the center is 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:
According to shell theorem, the gravitational field inside a uniform spherical shell is zero at all points. Thus, the force acting on the point mass is zero.
How much deep inside the earth (radius \(R\)) should a man go, so that his weight becomes one-fourth of that on the earth’s surface?
The acceleration due to gravity at depth \(d\) is \(g' = g \left(1 - \frac{d}{R}\right)\). Setting \(g' = g/4\), we get \(1 - \frac{d}{R} = \frac{1}{4}\) which gives \(d = \frac{3R}{4}\).
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of \(g\) on the earth’s surface would
Since \(g = \frac{GM}{R^2}\), differentiating gives \(\frac{dg}{g} = -2 \frac{dR}{R}\). A -1% change in R leads to a +2% change in g.
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.
Gravitational potential difference between a point on surface of planet and another point 10m above is 4J/kg. Considering gravitational field to be uniform, how much work is done in moving a mass of 2.0 kg from the surface to a point 5.0m above the surface?
Since the field is uniform, potential varies linearly with height. Potential difference at \(5.0\text{ m}\) is \(\Delta V = 4 \times \frac{5.0}{10} = 2\text{ J/kg}\. Work done is \(W = m \Delta V = 2.0 \times 2 = 4.0\text{ J}\).
Two identical particles of combined mass \(M\), placed in space with certain separation, are released. Interaction between the particles is only of gravitational in nature and there is no external force present. Acceleration of one particle with respect to the other when separation between them is \(R\), has a magnitude :
Each particle has mass \(m = M/2\). The force is \(F = \frac{G m^2}{R^2} = \frac{GM^2}{4R^2}\). Acceleration of each is \(a = \frac{F}{m} = \frac{GM}{2R^2}\). Relative acceleration is \(a_{\text{rel}} = 2a = \frac{GM}{R^2}\).
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 potential at the center of a shell of mass \(M\) and radius \(R\) is \(-\frac{GM}{R}\). By superposition, the total potential at the common center is \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).
If 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 the centers of any two touching spheres is \(2R\). The gravitational force between any two is \(F = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}\). The angle between the two forces acting on one sphere is \(60^\circ\). Net force is \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3}GM^2}{4R^2}\).