Acceleration Due to Gravity and its Variation Questions in English

(C) The variation in acceleration due to gravity at a height $h$ above the Earth's surface is given by $g' = g(1 - 2h/R)$.
Given $g' = 9 m s^{-2}$ and $h = R/20$,we have:
$9 = g(1 - 2(R/20)/R) = g(1 - 1/10) = g(9/10)$.
Therefore,$g = 10 m s^{-2}$.
The variation in acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g'' = g(1 - d/R)$.
For an equal depth $d = h = R/20$,we have:
$g'' = 10(1 - (R/20)/R) = 10(1 - 1/20) = 10(19/20) = 9.5 m s^{-2}$.

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by $g_d = g(1 - \frac{d}{R})$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given that the acceleration due to gravity is reduced by $40 \%$,the value at depth $d$ becomes $g_d = g - 0.40g = 0.60g$.
Substituting this into the formula: $0.60g = g(1 - \frac{d}{R})$.
Dividing both sides by $g$: $0.60 = 1 - \frac{d}{R}$.
Rearranging the terms: $\frac{d}{R} = 1 - 0.60 = 0.40$.
Therefore,$d = 0.40 \times R$.
Given $R = 6400 \ km$,we have $d = 0.40 \times 6400 \ km = 2560 \ km$.

(A) The acceleration due to gravity at height $h$ is given by $g_h = g \left( \frac{R}{R+h} \right)^2$. Given $h = 1600 \ km$ and $R = 6400 \ km$,we have $g_h = g \left( \frac{6400}{6400+1600} \right)^2 = g \left( \frac{6400}{8000} \right)^2 = g \left( \frac{4}{5} \right)^2 = 0.64g$.
The acceleration due to gravity at depth $d$ is given by $g_d = g \left( 1 - \frac{d}{R} \right)$.
According to the problem,$g_d = \frac{1}{2} g_h = \frac{1}{2} (0.64g) = 0.32g$.
Substituting this into the depth formula: $0.32g = g \left( 1 - \frac{d}{R} \right)$.
$0.32 = 1 - \frac{d}{R} \Rightarrow \frac{d}{R} = 1 - 0.32 = 0.68$.
$d = 0.68 \times 6400 \ km = 4352 \ km = 4.352 \times 10^6 \ m$.
Note: If using the approximation $g_h \approx g(1 - 2h/R)$,then $g_h = g(1 - 3200/6400) = 0.5g$. Then $g_d = 0.5 g_h = 0.25g$.
$0.25g = g(1 - d/R) \Rightarrow d/R = 0.75 \Rightarrow d = 0.75 \times 6400 \ km = 4800 \ km = 4.8 \times 10^6 \ m$.
Given the options,the approximation method is intended.

(C) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the earth's rotation and $R$ is the radius of the earth.
At the poles,the latitude $\lambda = 90^\circ$. Substituting this into the formula,we get $g' = g - \omega^2 R \cos^2(90^\circ) = g - 0 = g$.
Since the value of $g'$ at the poles is independent of the angular velocity $\omega$,the weight of a body $(w = mg')$ will not change if the earth stops rotating on its own axis.
Therefore,there will be no variation in the weight of our bodies at the poles.

(A) Given: Radius of Earth $(R_e)$ $= 6400 \,km$,Radius of Mars $(R_m)$ $= 3200 \,km$.
Mass of Earth $(M_e)$ $= 10 M_m$,where $M_m$ is the mass of Mars.
The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
The ratio of acceleration due to gravity on Mars $(g_m)$ to that on Earth $(g_e)$ is:
$\frac{g_m}{g_e} = \frac{G M_m / R_m^2}{G M_e / R_e^2} = \frac{M_m}{M_e} \times \left(\frac{R_e}{R_m}\right)^2$.
Substituting the given values:
$\frac{g_m}{g_e} = \frac{1}{10} \times \left(\frac{6400}{3200}\right)^2 = \frac{1}{10} \times (2)^2 = \frac{4}{10} = \frac{2}{5}$.
The weight of an object is $W = mg$.
Given weight on Earth $W_e = m g_e = 200 \,N$.
Weight on Mars $W_m = m g_m = m \left(\frac{2}{5} g_e\right) = \frac{2}{5} W_e$.
$W_m = \frac{2}{5} \times 200 \,N = 80 \,N$.

(B) Given that,the angle of latitude is $\lambda = 45^{\circ}$.
The effective acceleration due to gravity at a latitude $\lambda$ is given by $g_{\lambda} = g - \omega^2 R \cos^2 \lambda$.
The apparent weight of a man of mass $m$ at this point is $w = m g_{\lambda} = m(g - \omega^2 R \cos^2 \lambda)$.
According to the question,the man feels weightlessness,so $w = 0$.
Therefore,$m(g - \omega^2 R \cos^2 45^{\circ}) = 0$.
Since $m \neq 0$,we have $g - \omega^2 R (\frac{1}{\sqrt{2}})^2 = 0$,which simplifies to $g - \frac{\omega^2 R}{2} = 0$.
This gives $\omega^2 = \frac{2g}{R}$,or $\omega = \sqrt{\frac{2g}{R}}$.
The duration of a day (time period $T$) is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{2g/R}} = 2\pi \sqrt{\frac{R}{2g}} = \pi \sqrt{\frac{2R}{g}}$.

(B) The acceleration due to gravity $g'$ at an altitude $h$ above the Earth's surface is given by the formula:
$g' = g \left( 1 + \frac{h}{R_e} \right)^{-2}$
Using the binomial theorem for $h \ll R_e$,we can approximate this as:
$g' \approx g \left( 1 - \frac{2h}{R_e} \right)$
From this expression,it is clear that as the altitude $h$ increases,the term $\frac{2h}{R_e}$ increases,causing the value of $g'$ to decrease.
Therefore,the acceleration due to gravity decreases with altitude.

(B) Let $g$ be the acceleration due to gravity on the surface of the earth and $R_e$ be the radius of the earth.
The acceleration due to gravity at a height $h$ is given by the formula:
$g_h = \frac{g}{(1 + \frac{h}{R_e})^2}$
According to the problem, $g_h = \frac{g}{2}$.
Substituting this into the equation:
$\frac{g}{2} = \frac{g}{(1 + \frac{h}{R_e})^2}$
$(1 + \frac{h}{R_e})^2 = 2$
$1 + \frac{h}{R_e} = \sqrt{2}$
$h = (\sqrt{2} - 1) R_e$
Using $R_e \approx 6400 \,km$ and $\sqrt{2} \approx 1.414$:
$h = (1.414 - 1) \times 6400 \,km$
$h = 0.414 \times 6400 \,km$
$h = 2649.6 \,km$
Rounding to the nearest provided option, the height is approximately $2625 \,km$.

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$,where $R_e$ is the radius of the Earth.
The acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g_d = g(1 - \frac{d}{R_e})$.
According to the problem,$g_h = g_d$.
Equating the two expressions:
$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$
Canceling $g$ from both sides:
$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$
Subtracting $1$ from both sides:
$-\frac{2h}{R_e} = -\frac{d}{R_e}$
Multiplying by $-R_e$:
$2h = d$ or $d = 2h$.

(A) The acceleration due to gravity $g^{\prime}$ at a latitude $\lambda$ on the surface of the Earth is given by the formula:
$g^{\prime} = g - \omega^2 R_e \cos^2 \lambda$
where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the Earth,and $R_e$ is the radius of the Earth.
At the poles,the latitude $\lambda = 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the expression becomes:
$g^{\prime} = g - \omega^2 R_e (0)^2 = g$
At the equator,$\lambda = 0^{\circ}$,so $\cos 0^{\circ} = 1$,which gives $g^{\prime} = g - \omega^2 R_e$,which is the minimum value.
Therefore,the value of acceleration due to gravity is maximum at the poles.

(C) The acceleration due to gravity at a depth $d$ below the earth's surface is given by the formula: $g_d = g(1 - \frac{d}{R})$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given that the depth $d = \frac{R}{2}$,we substitute this into the formula:
$g_d = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
Since weight $w = mg$,the new weight $w'$ at depth $d$ is $w' = m g_d = m(\frac{g}{2}) = \frac{mg}{2} = \frac{w}{2}$.
Therefore,the weight of the body at a depth half way to the centre of the earth is $\frac{w}{2}$.

(B) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth (where $r < R_e$) is given by $F = -\frac{GMmr}{R_e^3}$.
Since $F \propto -r$,the force is a restoring force proportional to the displacement from the center,which is the condition for Simple Harmonic Motion $(SHM)$. Thus,Assertion $(A)$ is true.
The Reason $(R)$ states Newton's Law of Gravitation,which is also a true statement $(F \propto 1/r^2)$.
However,the motion inside the Earth is governed by the effective gravitational field inside a sphere,where the force is proportional to $r$,not $1/r^2$. Therefore,$(R)$ is not the correct explanation for $(A)$.

(A) At the centre of the Earth, the gravitational acceleration $g$ is zero.
Since the weight $w$ of an object is defined as the product of its mass $m$ and the gravitational acceleration $g$, we have $w = m \times g$.
Substituting the given values, $w = 10 \,kg \times 0 \,m/s^2 = 0 \,N$.
Therefore, the weight of the point mass at the centre of the Earth is zero.

(C) The apparent acceleration due to gravity $g^{\prime}$ at the equator due to the rotation of the Earth is given by the formula:
$g^{\prime} = g_0 - \omega^2 R$
Given that the apparent $g^{\prime}$ is $(1/6)$th of its original value $g_0$,we have:
$g^{\prime} = \frac{g_0}{6}$
Substituting this into the equation:
$\frac{g_0}{6} = g_0 - \omega^2 R$
Rearranging the terms to solve for $\omega^2 R$:
$\omega^2 R = g_0 - \frac{g_0}{6} = \frac{5}{6} g_0$
$\omega = \sqrt{\frac{5 g_0}{6 R}}$
Substituting the values $g_0 = 9.8 \ m/s^2$ and $R = 6.4 \times 10^6 \ m$:
$\omega = \sqrt{\frac{5 \times 9.8}{6 \times 6.4 \times 10^6}}$
$\omega = \sqrt{\frac{49}{38.4 \times 10^6}} = \sqrt{1.276 \times 10^{-6}}$
$\omega \approx 1.13 \times 10^{-3} \ rad \ s^{-1}$
Thus,the angular speed is approximately $1.14 \times 10^{-3} \ rad \ s^{-1}$.

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
For Earth,$g_e = \frac{GM}{R^2}$.
For the new planet,the mass $M' = 100M$ and the radius $R' = 10R$.
Therefore,the acceleration due to gravity on the new planet is $g_p = \frac{G(100M)}{(10R)^2} = \frac{100GM}{100R^2} = \frac{GM}{R^2} = g_e$.
Since the height $h$ is the same and the acceleration due to gravity $g$ is the same,the time taken to reach the ground is given by $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are identical for both,the time taken $t$ remains the same.

(C) The gravitational force on a body of mass $m$ at the surface of the earth is $F_0 = \frac{G M_e m}{R^2}$.
At a height $h$,the force is $F_h = \frac{G M_e m}{(R+h)^2}$.
The error in weighing is the difference in forces: $\Delta F = F_0 - F_h = \frac{G M_e m}{R^2} - \frac{G M_e m}{(R+h)^2}$.
$\Delta F = \frac{G M_e m}{R^2} \left[ 1 - (1 + \frac{h}{R})^{-2} \right]$.
Using binomial expansion $(1+x)^n \approx 1+nx$ for $h \ll R$,we get $(1 + \frac{h}{R})^{-2} \approx 1 - \frac{2h}{R}$.
So,$\Delta F \approx \frac{G M_e m}{R^2} \left[ 1 - (1 - \frac{2h}{R}) \right] = \frac{G M_e m}{R^2} \left( \frac{2h}{R} \right) = \frac{2 G M_e m h}{R^3}$.
Since the density $\rho = \frac{M_e}{\frac{4}{3} \pi R^3}$,we have $M_e = \frac{4}{3} \pi R^3 \rho$.
Substituting $M_e$ into the expression: $\Delta F = \frac{2 G m h}{R^3} \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{8}{3} \pi \rho G m h$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
At height $h$ above the surface, the acceleration due to gravity is $g' = \frac{g}{(1 + \frac{h}{R})^2}$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g / (1 + \frac{h}{R})^2}} = T(1 + \frac{h}{R})$.
Given $T' = 2T$, we have $2T = T(1 + \frac{h}{R})$.
$2 = 1 + \frac{h}{R} \Rightarrow \frac{h}{R} = 1$.
Therefore, $h = R = 6400 \text{ km}$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g'}}$.
Here,$g'$ is the acceleration due to gravity at a height $h$ from the surface of the earth,given by $g' = g (\frac{R_E}{R_E + h} )^2$.
Therefore,$T \propto \frac{1}{\sqrt{g'}} \propto \frac{R_E + h}{R_E}$.
At height $h_1 = 2 R_E$,the time period is $T_1 \propto \frac{R_E + 2 R_E}{R_E} = \frac{3 R_E}{R_E} = 3$.
At height $h_2 = 3 R_E$,the time period is $T_2 \propto \frac{R_E + 3 R_E}{R_E} = \frac{4 R_E}{R_E} = 4$.
The ratio of the time periods is $\frac{T_1}{T_2} = \frac{3}{4}$ or $3: 4$.

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given $R = 6400 \ km$.
For $h_1 = 1280 \ km$:
$g_1 = g \left( \frac{6400}{6400 + 1280} \right)^2 = g \left( \frac{6400}{7680} \right)^2 = g \left( \frac{5}{6} \right)^2 = \frac{25}{36} g$.
For $h_2 = 3200 \ km$:
$g_2 = g \left( \frac{6400}{6400 + 3200} \right)^2 = g \left( \frac{6400}{9600} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9} g$.
Now,the ratio $\frac{g_1}{g_2}$ is:
$\frac{g_1}{g_2} = \frac{25/36 g}{4/9 g} = \frac{25}{36} \times \frac{9}{4} = \frac{25}{16}$.
Thus,the ratio is $25:16$.

(D) The acceleration due to gravity at an altitude $h$ is given by $g_h = \frac{GM}{(R+h)^2}$.
Since $g_h$ is inversely proportional to the square of the distance from the center,$g_h$ decreases as altitude $h$ increases.
Acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,which clearly depends on the mass of the Earth $M$.
$A$ geostationary satellite must have a time period of exactly $24 \ h$ to remain stationary relative to the Earth's surface.
The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - \frac{d}{R}) = g(\frac{R-d}{R})$.
As depth $d$ increases,the term $(\frac{R-d}{R})$ decreases,so $g_d$ decreases with increasing depth.
Therefore,option $D$ is the correct statement.

(B) The variation of acceleration due to gravity $(g)$ with distance $(r)$ from the centre of the earth is given by:
$1$. Inside the earth $(r < R_e)$: The acceleration due to gravity is $g' = \frac{GM r}{R_e^3}$. Since $G, M, R_e$ are constants,$g' \propto r$. This represents a straight line passing through the origin.
$2$. Outside the earth $(r \geq R_e)$: The acceleration due to gravity is $g' = \frac{GM}{r^2}$. Thus,$g' \propto \frac{1}{r^2}$. This represents a rectangular hyperbola.
At the surface $(r = R_e)$,the value of $g$ is maximum. Combining these two,the graph shows a linear increase up to $r = R_e$ and then a decrease following an inverse-square law. Therefore,the correct graph is the one showing a linear rise to a peak at $R_e$ followed by a curve.

(C) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.
Taking the logarithmic differentiation,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $1 \%$,we have $\frac{\Delta R}{R} = -0.01$.
Substituting this value,we get $\frac{\Delta g}{g} = -2 \times (-0.01) = 0.02$.
Therefore,the percentage change in $g$ is $0.02 \times 100 = 2 \%$.
Since the value is positive,the acceleration due to gravity increases by $2 \%$.

(B) The gravitational acceleration at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R})$,where $R$ is the radius of the Earth.
The gravitational acceleration at a depth $d$ below the Earth's surface is given by $g_d = g(1 - \frac{d}{R})$.
Given that the gravitational accelerations are the same,we equate the two expressions:
$g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$
Canceling $g$ from both sides:
$1 - \frac{2h}{R} = 1 - \frac{d}{R}$
Subtracting $1$ from both sides:
$-\frac{2h}{R} = -\frac{d}{R}$
Multiplying by $-R$:
$2h = d$
Therefore,the ratio of height to depth is:
$\frac{h}{d} = \frac{1}{2} = 0.5$

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$, where $\rho$ is the mean density and $R$ is the radius of the planet.
From this relation, we see that $g \propto \rho R$.
Given for the planet $(1)$ and the earth $(2)$:
Ratio of radii: $\frac{R_1}{R_2} = \frac{1}{2}$
Ratio of densities: $\frac{\rho_1}{\rho_2} = \frac{4}{1}$
Acceleration due to gravity on earth: $g_2 = 9.8 \,ms^{-2}$.
Using the proportionality $g_1 / g_2 = (\rho_1 / \rho_2) \times (R_1 / R_2)$:
$\frac{g_1}{g_2} = \frac{4}{1} \times \frac{1}{2} = 2$.
Therefore, $g_1 = 2 \times g_2 = 2 \times 9.8 \,ms^{-2} = 19.6 \,ms^{-2}$.

(D) The Earth's internal structure consists of layers with increasing density and pressure as one moves towards the core.
At the center of the Earth (the inner core),the gravitational compression from the overlying layers is at its maximum,resulting in the highest pressure.
Due to the intense pressure and the presence of radioactive decay and residual heat from planetary formation,the temperature is also highest at the center.
Therefore,the center of the Earth exhibits the highest values for temperature,density,and pressure.

(B) The mass of a body is defined as the quantity of matter contained in it and is an intrinsic property of the object.
Mass remains constant regardless of the location of the body in the universe,whether it is on the surface of the earth,at a depth $d$ below the surface,or at an altitude $h$ above the surface.
While the weight of the body changes due to the variation in the acceleration due to gravity $(g)$,the mass of the body does not change.
Therefore,the change in mass is zero,and it remains constant.

(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left(1 + \frac{h}{R}\right)^{-2}$.
Given that the acceleration due to gravity at height $h$ is $1 \%$ of its value at the surface,we have $g' = \frac{1}{100} g$.
Substituting this into the formula:
$\frac{g}{100} = g \left(1 + \frac{h}{R}\right)^{-2}$
$\frac{1}{100} = \left(1 + \frac{h}{R}\right)^{-2}$
Taking the square root on both sides:
$\frac{1}{10} = \left(1 + \frac{h}{R}\right)^{-1}$
$1 + \frac{h}{R} = 10$
$\frac{h}{R} = 9$
$h = 9 R$.
Therefore,the height is $9 R$.

(B) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - \frac{d}{R})$.
The acceleration due to gravity at a height $h$ is given by $g_h = g(1 - \frac{2h}{R})$.
We are moving from depth $d = 16 \text{ km}$ to height $h = 16 \text{ km}$.
The change in $g$ is $\Delta g = g_h - g_d = g(1 - \frac{2h}{R}) - g(1 - \frac{d}{R})$.
Substituting the values: $\Delta g = g(1 - \frac{2 \times 16}{6400}) - g(1 - \frac{16}{6400}) = g(1 - \frac{32}{6400} - 1 + \frac{16}{6400}) = g(-\frac{16}{6400}) = -\frac{g}{400}$.
The percentage change $\alpha$ is given by $|\frac{\Delta g}{g}| \times 100 = |-\frac{1}{400}| \times 100 = 0.25 \%$.
Thus,the value of $\alpha$ is $0.25$.

(B) The formula for acceleration due to gravity at a height $h$ above the earth's surface is given by: $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = g/9$,we substitute this into the equation:
$\frac{g}{9} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$:
$\frac{1}{9} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides:
$\frac{1}{3} = \frac{R}{R+h}$.
Cross-multiplying gives:
$R + h = 3R$.
Therefore,$h = 3R - R = 2R$.