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 \(2%\).
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 ?
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.
The acceleration due to gravity increases by \(0.5\%\) when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles?
The time period of a seconds pendulum at the poles is \(T_p = 2\text{ s}\). Since \(T propto g^{-1/2}\), we have \(frac{T_e}{T_p} = sqrt{frac{g_p}{g_e}} = sqrt{1 + 0.005} approx 1 + 0.0025\). This gives \(T_e approx 2 times 1.0025 = 2.005\text{ s}\).
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 \(m \cdot dg = \frac{8}{3} \pi G \rho m h\).
Assertion : If rotation of earth about its own axis is suddenly stopped then acceleration due to gravity will increase at all places on the earth (except poles).
Reason : At height \(h\) from the surface of earth, acceleration due to gravity is \(g_h = g \left(1 – \frac{2h}{R_e}\right)\) (If \(h \ll R_e\)) [\(R_e \rightarrow\) radius of earth]
The effective gravity is \(g' = g - \omega^2 R \cos^2\lambda\). If rotation stops (\(\omega = 0\)), \(g'\) increases everywhere except the poles (where \(\lambda = 90^\circ\)). Thus Assertion is true. The Reason is also a true independent formula for gravity at a height, but not the explanation.
The weight of a body at a distance \(r\) (\(r > R\)) from the centre of earth is \(W\). The weight at a distance \(2r\) from the centre of earth is : (\(R \rightarrow\) Radius of earth)
Outside the earth, weight \(W \propto \frac{1}{r^2}\). When distance is doubled to \(2r\), the weight becomes \(\frac{1}{2^2} = \frac{1}{4}\) of its initial value, i.e., \(\frac{W}{4}\).