The 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.
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.
Suppose the gravitational force varies inversely as the \(n^{\text{th}}\) power of distance. Then, the time period of a planet in circular orbit of radius \(R\) around the sun will be proportional to
The centripetal force is provided by the gravitational force: \(m \omega^2 R = \frac{k}{R^n} ⇒ \omega^2 \propto \frac{1}{R^{n+1}}\). Since \(T = \frac{2\pi}{\omega}\), we get \(T^2 \propto R^{n+1} ⇒ T \propto R^{\frac{n+1}{2}}\).
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.
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.
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 \(a_{\text{rel}} = 2a = \frac{GM}{R^2}\).
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 \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3}GM^2}{4R^2}\).
Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \(r^{-5/2}\), then the square of the time period will be proportional to
The centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).
Assertion (A): The gravitational force between two finite bodies is necessarily along the line joining their centre of mass.
Reason (R): The gravitational force between two particles is not central.
Assertion (A) is false.
The gravitational force between two finite bodies is generally not directed along the line joining their centers of mass.
Reason (R) is also false. The gravitational force between two point particles is always central, acting along the line connecting them. Therefore, both (A) and (R) are false.