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.
If the earth stops rotating about its axis, the acceleration due to gravity will remain unchanged at
The effective gravity is \(g' = g - \omega^2 R \cos^2 \theta\). At the poles, \(\theta = 90^\circ\), so \(g' = g\), which is independent of the earth's angular velocity.
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 \(F_{\text{net}} = \sqrt{3}F = \frac{\sqrt{3}GM^2}{4R^2}\).
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.
At certain height \(h\) from surface of earth the value of \(g\) become \(6.25%\) of its value at earth surface. The ratio of \(\frac{h}{R}\) is. (\(R\) is radius of earth)
The variation of gravity with height is \(g' = g \left(\frac{R}{R+h}\right)^2\). Given \(g' = 0.0625 g = \frac{1}{16} g\), we get \(\frac{R}{R+h} = \frac{1}{4} ⇒ R+h = 4R ⇒ h = 3R ⇒ \frac{h}{R} = 3\).
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 \(g' = g \left(1 - \frac{d}{R}\right)\). For \(g' = \frac{g}{4}\), we have \(1 - \frac{d}{R} = \frac{1}{4} ⇒ \frac{d}{R} = \frac{3}{4} ⇒ d = \frac{3R}{4}\).
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 \(g = \frac{GM}{R^2}\), for a small percentage change: \(\frac{\Delta g}{g} \times 100 = -2 \frac{\Delta R}{R} \times 100\). Since \(R\) decreases by \(1%\), \(g\) increases by \(2%\).