The depth at which the value of acceleration due to gravity becomes 1/n times the value at the surface is (R be the radius of the earth) :
The depth at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the surface value is:
\[
d = \left( 1 - \frac{1}{n} \right) R
\]
where \( R \) is the radius of the Earth.
The height at which the weight of a body becomes 1/16th, its weight on the surface of earth (radius R), is :
The weight of a body at height \( h \) is given by:
\[
W_h = W_0 \left( \frac{R}{R + h} \right)^2.
\]
To find \( h \) when \( W_h = \frac{1}{16} W_0 \):
\[
\frac{1}{16} = \left( \frac{R}{R + h} \right)^2.
\]
Taking the square root:
\[
\frac{1}{4} = \frac{R}{R + h}.
\]
Cross-multiplying gives:
\[
R + h = 4R ; h = 3R.
\]
Thus, the height is \( 3R \).
The height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth is: (Assume radius of the earth is R) :
The height \( h \) at which gravity decreases by 36% (to 64% of its surface value) is found using:
\[
g_h = \frac{gR^2}{(R + h)^2}.
\]
Setting \( g_h = 0.64g \):
\[
0.64 = \frac{R^2}{(R + h)^2}.
\]
Taking the square root:
\[
0.8 = \frac{R}{R + h}.
\]
Cross-multiplying gives:
\[
0.8(R + h) = R \implies 0.8h = 0.2R \implies h = 0.25R=R/4
\]
Thus, the height is R/4.
The height at which the weight of a body becomes 1/16th, its height from the surface of earth (radius R), is :
To find the height \( h \) at which the weight of a body becomes \( \frac{1}{16} \) of its weight on the surface of the Earth, use the formula:
\[
W_h = W_0 \left( \frac{R}{R + h} \right)^2.
\]
Setting \( W_h = \frac{1}{16} W_0 \):
\[
\frac{1}{16} = \left( \frac{R}{R + h} \right)^2.
\]
Taking the square root:
\[
\frac{1}{4} = \frac{R}{R + h}.
\]
Cross-multiplying gives:
\[
R + h = 4R ; h = 4R - R = 3R.
\]
Thus, the height is \( 3R \).
If the rotational motion of earth is suddenly stopped then which one is correct :
If the Earth's rotational motion were suddenly stopped, the value of \( g \) would effectively increase everywhere except at the poles due to the following reasons:
1. Centrifugal Force: The Earth's rotation creates a centrifugal force that slightly counteracts the force of gravity. At the equator, this effect is the greatest, reducing the apparent weight and thus the effective value of \( g \). When rotation stops, this centrifugal force disappears.
2. At the Poles: At the poles, there is no centrifugal force due to rotation since they are the axis of rotation. Therefore, the value of \( g \) would remain unchanged at the poles.
Consequently, \( g \) would increase at all places on Earth except at the poles, where it would stay the same.
The mass of the moon is about 1.2% of the mass of the earth. compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth :
From Newton's Third law of motion force applied by object A on B and Force applied by B on A will have equal magnitude.
The average density of the earth :
The average density of the Earth (\( \rho \)) is directly proportional to the acceleration due to gravity (\( g \)) at the surface due to the relationship expressed by the formula:
\[
g = \frac{G M}{R^2},
\]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.
The mass \( M \) can be expressed in terms of density and volume:
\[
M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right).
\]
Substituting this into the equation for \( g \):
\[
g = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4\pi G \rho R}{3}.
\]
This shows that \( g \) is directly proportional to \( \rho \) (as \( R \) and \( G \) are constants), meaning that as the average density of the Earth increases, the value of \( g \) also increases.
A body released from a height h takes time t to reach earth’s surface. The time taken by the same body released from the same height to reach the moon’s surface is :
The time taken for an object to fall freely from a height \( h \) is given by the equation:
\[
h = \frac{1}{2} g t^2,
\]
where \( g \) is the acceleration due to gravity.
1. For Earth:
\[
h = \frac{1}{2} g_E t^2 ; t = \sqrt{\frac{2h}{g_E}}.
\]
2. For the Moon, where \( g_M \approx \frac{1}{6} g_E \):
\[
h = \frac{1}{2} g_M t_m^2 ; t_m = \sqrt{\frac{2h}{g_M}} = \sqrt{\frac{2h}{\frac{1}{6} g_E}} = \sqrt{12 \cdot \frac{2h}{g_E}}.
\]
This can be simplified using the time \( t \) from Earth:
\[
t_m = \sqrt{6} \cdot t = \sqrt{6} \cdot t.
\]
Thus, the time taken by the same body to reach the Moon's surface is approximately \( \sqrt{6} t \).
If \(R\) is the radius of the earth and \(g\) is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be
Acceleration due to gravity is \(g = \frac{GM}{R^2}\). Since mass \(M = \rho \times \frac{4}{3}\pi R^3\), we get \(g = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi \rho GR\). Rearranging gives \(\rho = \frac{3g}{4\pi RG}\).
Consider earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The value of g measured by:
Acceleration due to gravity decreases with depth as \( g_d = g(1 - \frac{d}{R}) \) and with height as \( g_h = g(1 - \frac{2h}{R}) \). Since the formulas and rates of decrease are different, they decrease at different rates.