The diameter of two planets are in the ratio 4:1 and their mean densities in the ratio 1:2 the acceleration due to gravity on the planets will be in ratio :
\[ g=\frac{GM}{R^{2}}= \frac{G\times \rho\frac{4}{3}\pi R^{3}}{R^{2}}=\frac{4G\rho\pi R}{3} \]
\[ \frac{g_{1}}{g_{2}}= \frac{R_{1}.\rho_{1}}{R_{2}.\rho_{2}}= \frac{4\times 1}{1\times 2}= \frac{2}{1} \]
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 \( 3R \).
Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g’, then :
The acceleration due to gravity \( g' \) on the new planet can be calculated using the formula:
\[
g' = \frac{GM}{R^2}.
\]
For the new planet:
- Its radius \( R' = 3R \) (3 times the radius of Earth).
- Its mass \( M' \) is given by \( M' = \text{density} \times \text{volume} = \rho \times \frac{4}{3} \pi (R')^3 = \rho \times \frac{4}{3} \pi (3R)^3 = 27 \times \rho \times \frac{4}{3} \pi R^3 = 27M \) (mass is 27 times that of Earth).
Substituting into the formula:
\[
g' = \frac{G(27M)}{(3R)^2} = \frac{27GM}{9R^2} = 3g.
\]
Thus, \( g' = 3g \).
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 \( 3R \).
A spherical planet has a mass Mp and diameter Dp. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to :
The acceleration due to gravity \( g \) on the surface of a spherical planet can be calculated using the formula:
\[
g = \frac{GM_p}{R_p^2},
\]
where:
- \( G \) is the gravitational constant,
- \( M_p \) is the mass of the planet,
- \( R_p \) is the radius of the planet.
Given that the diameter \( D_p = 2R_p \), we can express the radius as \( R_p = \frac{D_p}{2} \).
Substituting this into the formula gives:
\[
g = \frac{GM_p}{\left(\frac{D_p}{2}\right)^2} = \frac{GM_p}{\frac{D_p^2}{4}} = \frac{4GM_p}{D_p^2}.
\]
Thus, the acceleration due to gravity experienced by the particle near the surface of the planet is:
\[
g = \frac{4GM_p}{D_p^2}.
\]
The change in the value of g at a height h above the surface of the earth is the same as at a depth d below the earth when both d and h are much smaller than the radius of the earth, which one of the following is correct ?
The change in the value of \( g \) at height \( h \) above the surface of the Earth and at depth \( d \) below the surface can be expressed using the following formulas:
1. **At height \( h \)**:
\[
g_h = g \left(1 - \frac{2h}{R}\right) \quad \text{(for small } h\text{)}
\]
The change in \( g \) is:
\[
\Delta g_h = g - g_h = g \frac{2h}{R} = \frac{2gh}{R}.
\]
2. **At depth \( d \)**:
\[
g_d = g \left(1 - \frac{d}{R}\right) \quad \text{(for small } d\text{)}
\]
The change in \( g \) is:
\[
\Delta g_d = g - g_d = g \frac{d}{R}.
\]
Setting the changes equal gives:
\[
\frac{2gh}{R} = \frac{g d}{R}.
\]
Cancelling \( g \) and \( R \) (assuming they are non-zero):
\[
2h = d.
\]
Thus, the relation between depth \( d \) and height \( h \) is:
\[
d = 2h.
\]
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.