Gravitation Practice Questions

At height r from center of earth. orbital velocity = $\sqrt{{GM}/r}$ ∴ By energy conservation HE of 'm' + $(-{GMm}/r)$ = 0 + 0 (At infinity, PE = KE = 0) \Rightarrow KE of 'm' = ${GMm}/r$ = $(\sqrt{{GM}/r})^2$ m = $mv^2$

As $T=2 \pi \sqrt{\ \frac{r^{3}}{G m}}$ $\Rightarrow T=2 \pi \times \sqrt{ \frac{\ (2 \\times 10^{6}\ )^{3}}{6.67 \times 10^{-11} \times 8 \\times 10^{22}}}$ ${T}=\ \frac{2 \pi \times 2871 \\times 10^{6}}{23 \\times 10^{5}}$ $=7840 \ {sec}$ $\Rightarrow$ No. of revolutions in 24 hrs $\ \frac{24 \\times 60 \\times 60}{7840}=11.02$

$E= \frac{G M m}{R}$ $\Rightarrow E= \frac{G \rho (\ \frac{4}{3} \pi R^{3} ) m}{R}$ $\Rightarrow E \propto R^{2}$ $\frac{E_{2}}{E_{1}}= ( \frac{R_{2}}{R_{1}} )^{2}$ $\Rightarrow E_{2}= ( \frac{1}{4} )^{2} E \Rightarrow \frac{E}{16}$

Orbital velocity V = $\sqrt{{GMe}/r}$ $T_A$ = $1/2 m_A V_A^2$ $T_B$ = $1/2 m_B V_B^2$ \Rightarrow $T_A/T_B$ = $\frac{m\times {Gm}/R}{2m\times \frac{Gm}{2R}}$ \Rightarrow $T_A/T_B$ = 1

$v_0$ = $\sqrt{g(R+h)}$ ~ $\sqrt{gR}$ $v_e$ = $\sqrt{2g(R+h)}$ ~ $\sqrt{2gR}$ Δv = $v_e-v_0$ = $(\sqrt2-1)\sqrt{gR}$

dm = (A + B$x^2$) dx = Gm $[-A/x+Bx]^{a+L}_a$ Gm $[A(1/{a}-1/{a+L})+BL]$

The acceleration due to gravity at an altitude $h$ is given by the formula $g_h = g_s \left( \frac{R}{R+h} \right)^2$.
Given $g_h = 4.9 \, \text{m/s}^2$ and $g_s = 9.8 \, \text{m/s}^2$, we have $\frac{4.9}{9.8} = \left( \frac{R}{R+h} \right)^2$, which simplifies to $\frac{1}{2} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides gives $\frac{1}{\sqrt{2}} = \frac{R}{R+h}$, which implies $R+h = R\sqrt{2}$.
Solving for $h$, we get $h = R(\sqrt{2} - 1)$.
Substituting $R = 6.4 \times 10^6 \, \text{m}$ and $\sqrt{2} \approx 1.414$, we calculate $h \approx 6.4 \times 10^6 \times (1.414 - 1) \approx 2.65 \times 10^6 \, \text{m}$.

Field at $P= \frac{G M}{(3 a)^{2}}+ \frac{G(2 M)}{(3 a)^{2}}$ Field $= \frac{ G M}{3 a^{2}}$

$U_{\text{surface}}$ + $E_1$ = $U_h$ KE of satellite is zero at earth surface & atheight h - ${GM_em}/R_e + E_1$ = - ${GM_e m}/(Re+h)$ $E_1$ = $GM_em$ $(1/R_e - 1/{R_e + h})$ $E_1$ = ${GM_e m}/(R_e +h)$ × $h/R_e$ Gravitational attraction $F_G$ = $ma_c$ = ${mv^2}/(R_e+h)$ $E_2$ \Rightarrow ${mv^2}/(R_e+h)$ = ${GM_e m}/(R_e+h)^2$ $mv^2$ = $\frac{GM_e}{2(R_e+h)}$ $E_1$ = $E_2$ $h/R_e$ = $1/2$ \Rightarrow h = $R_e/2$ = 3200 km

The tidal force difference $\Delta F$ for a celestial body of mass $M$ at distance $R$ from Earth (radius $R_e$) is given by the differential of the gravitational force:
$\Delta F = \frac{GMm_e}{(R-R_e)^2} - \frac{GMm_e}{(R+R_e)^2} \approx \frac{4GM m_e R_e}{R^3}$
Thus, the ratio $\frac{\Delta F_1}{\Delta F_2}$ is given by $\frac{M_m}{M_s} \cdot \left(\frac{R_s}{R_m}\right)^3$.
Substituting the given values:
$\frac{\Delta F_1}{\Delta F_2} = \frac{8 \times 10^{22}}{2 \times 10^{30}} \cdot \left(\frac{150 \times 10^6}{0.4 \times 10^6}\right)^3 = (4 \times 10^{-8}) \cdot (375)^3$
$\frac{\Delta F_1}{\Delta F_2} = 4 \times 10^{-8} \cdot 52.73 \times 10^6 \approx 2.109$
The value closest to $2.109$ is 2.

$F\propto 1/R^n$ $MR{4\pi ^2}/T^2\propto 1/R^n$ $T\propto R^{{n+1}/2}$

The effective acceleration due to gravity on Earth is given by $g_{\text{eff}} = g - \omega^2 R \cos^2 \lambda$, where $\omega$ is the angular velocity, $R$ is Earth's radius, and $\lambda$ is the latitude.
The weight of an object is expressed as $W = m g_{\text{eff}} = m(g - \omega^2 R \cos^2 \lambda)$.
When $\omega$ increases, the term $\omega^2 R \cos^2 \lambda$ increases, causing $g_{\text{eff}}$ and thus $W$ to decrease for all latitudes except at the poles.
At the poles, $\lambda = 90^\circ$, so $\cos 90^\circ = 0$, meaning the centrifugal term vanishes and the weight remains unchanged regardless of $\omega$.
Therefore, weight decreases everywhere on the Earth except at the poles.

The initial total energy of the mass $m$ in orbit of radius $R$ is $E_i = -\frac{GMm}{2R}$.
The final energy of the two equal masses $m/2$ at radii $R/2$ and $3R/2$ are:
$E_1 = -\frac{G M (m/2)}{2(R/2)} = -\frac{GMm}{2R}$ and $E_2 = -\frac{G M (m/2)}{2(3R/2)} = -\frac{GMm}{6R}$.
The total final energy is $E_f = E_1 + E_2 = -\frac{GMm}{2R} - \frac{GMm}{6R} = -\frac{4GMm}{6R} = -\frac{2GMm}{3R}$.
The difference in energy is $\Delta E = E_f - E_i = -\frac{2GMm}{3R} - \left(-\frac{GMm}{2R}\right) = \frac{-4GMm + 3GMm}{6R} = -\frac{GMm}{6R}$.

The mass $M(r)$ enclosed within a radius $r \le R$ is calculated as:
$M(r) = \int_{0}^{r} \rho(r') 4\pi (r')^2 dr' = \int_{0}^{r} \frac{k}{r'} 4\pi (r')^2 dr' = 4\pi k \int_{0}^{r} r' dr' = 2\pi k r^2$
The gravitational acceleration is $a(r) = \frac{G M(r)}{r^2}$. For $r \le R$:
$a(r) = \frac{G (2\pi k r^2)}{r^2} = 2\pi G k = \text{constant}$
For $r > R$, the enclosed mass is constant at $M(R) = 2\pi k R^2$, so:
$a(r) = \frac{G (2\pi k R^2)}{r^2} \propto \frac{1}{r^2}$
Thus, the acceleration is constant for $r \le R$ and decreases as $1/r^2$ for $r > R$.

$g'=g-\omega ^2R\text{cos}^2\theta$ ${3g}/4=g-\omega ^2R$ $w^2R=g/4$ $w=\sqrt{g/{4R}}$ $=\sqrt{10/{4\times 64\omega \times 10^3}}$ $=1/{2\times 8\times 100}$ $=1/1600=1/16\times 10^{-2}=0.6\times 10^{-3}$

$g={GMx}/R^3$ inside the Earth(straight line)$g={GM}/r^2$ outside the EarthWhere M is Mass of Earth

$F_G={GMm}/(R+h)^2$

Total area = AArea of sabc$={3A}/4$Area of sadc $=A/4$ $\frac{3A}{4t_1}=A/{4t_2}\Rightarrow t_1=3t_2$