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 \( \sqrt{6} t \).