Gravitation Practice Questions

The total mechanical energy $E$ of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$. Since $GM = gR_E^2$, the energy is $E = -\frac{mgR_E^2}{2r}$.
The energy required to change the orbit from $r_1 = 1.5R_E$ to $r_2 = 3R_E$ is:
$\Delta E = E_2 - E_1 = \left(-\frac{mgR_E^2}{2(3R_E)}\right) - \left(-\frac{mgR_E^2}{2(1.5R_E)}\right) = \frac{mgR_E}{2} \left(\frac{1}{1.5} - \frac{1}{3}\right)$
Substituting the values $m = 100 \text{ kg}$, $g = 10 \text{ m/s}^2$, and $R_E = 6 \times 10^6 \text{ m}$:
$\Delta E = \frac{100 \times 10 \times 6 \times 10^6}{2} \left(\frac{2}{3} - \frac{1}{3}\right) = (300 \times 10^6) \times \left(\frac{1}{3}\right) = 1000 \times 10^6 \text{ J}$
Comparing this to $\alpha \times 10^6 \text{ J}$, we get $\alpha = 1000$.

Let the gravitational force exerted by a mass $m$ at a corner on the center be represented as a vector $\vec{f} \propto m(\pm \hat{i} \pm \hat{j})$, where $k$ is a constant $\frac{G}{r^2}$.
For the first arrangement, the net force components are $F_{1x} = k(3+2-4-1) = 0$ and $F_{1y} = k(4+3-1-2) = 4k$, so $F_1 = 4k$.
In the second arrangement, swapping $3M$ and $4M$ at the top corners results in $F_{2x} = k(4+2-3-1) = 2k$ and $F_{2y} = k(3+4-1-2) = 4k$.
The magnitude $F_2 = \sqrt{(2k)^2 + (4k)^2} = k\sqrt{4+16} = 2\sqrt{5}k$.
The ratio is $\frac{F_1}{F_2} = \frac{4k}{2\sqrt{5}k} = \frac{2}{\sqrt{5}}$.
Comparing this to $\frac{\alpha}{\sqrt{5}}$, we find $\alpha = 2$.

The work done in rearranging the masses is equal to the change in gravitational potential energy: $W = U_f - U_i$.
The gravitational potential energy for three masses at the vertices of an equilateral triangle of side $a$ is given by $U = -G \left( \frac{m_1 m_2 + m_2 m_3 + m_3 m_1}{a} \right)$.
Calculating the sum of the products of the masses: $S = (200 \cdot 300) + (300 \cdot 400) + (400 \cdot 200) = 60,000 + 120,000 + 80,000 = 2.6 \times 10^5 \, \text{kg}^2$.
The work done is $W = G \cdot S \left( \frac{1}{a_1} - \frac{1}{a_2} \right)$, where $a_1 = 20 \, \text{m}$ and $a_2 = 25 \, \text{m}$.
Substituting the values: $W = (6.7 \times 10^{-11}) \cdot (2.6 \times 10^5) \cdot \left( \frac{1}{20} - \frac{1}{25} \right)$.
$W = (17.42 \times 10^{-6}) \cdot (0.05 - 0.04) = 17.42 \times 10^{-8} \, \text{J} = 1.74 \times 10^{-7} \, \text{J}$.

For a satellite orbiting very close to the Earth's surface ($r \approx R_e$), the gravitational force provides the centripetal force: $\frac{mv^2}{R_e} = \frac{GMm}{R_e^2}$. The time period is $T = \frac{2\pi R_e}{v} = 2\pi \sqrt{\frac{R_e^3}{GM}}$. Using $g = \frac{GM}{R_e^2}$, we substitute $GM = gR_e^2$ into the expression to get $T = 2\pi \sqrt{\frac{R_e^3}{gR_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$, confirming Statement II is true. To verify Statement I, substitute the mass of Earth $M = \rho \left( \frac{4}{3}\pi R_e^3 \right)$ into the time period formula: $T = 2\pi \sqrt{\frac{R_e^3}{G (\frac{4}{3}\pi R_e^3 \rho)}} = \sqrt{\frac{3\pi}{G\rho}}$. Since $T$ depends on the density $\rho$, Statement I is also true.

From Kepler's $2^{\;\text{nd }\;}$ law$\;\frac{d A}{d t}=\;\frac{L}{2 m}=\;\text{ constant }\;$and angular momentum for central force is constant.so, both $(A)$ and $(R)$ are correct and $(R)$ explain $(A)$

$T^2 \propto r^3$ from Kepler's $3^{\;\text{rd }\;}$ law$\;\;\frac{27^2}{T^2}=\;\frac{r^3}{(r / g)^3}$ $\;\;\frac{27^2}{9^3}=T^2$ $\;\;\frac{27 \times 27}{9 \times 9 \times 9}=T^2 \Rightarrow T=1 \;\text{ day }\;$

(A) $F=\;\frac{G m^2}{r^2} \equiv G=\;\frac{F r^2}{m^2} \equiv \;\frac{M L T^{-2} L^2}{M^2}=M^{-1} L^3 T^{-2}$ $A \longrightarrow {\text{IV}}$ (B) $\;G . P E \equiv \;\text{ Energy }\; \equiv M L^2 T^{-2}$ $\;B \longrightarrow \;\text{ III }\;$ (C) $G . P=\;\frac{m g h}{m}=g h \equiv V^2 \equiv L^2 T^{-2}$ $C \longrightarrow I I$ (D) $\;g \equiv L T^{-2}$ $\;D \longrightarrow I$

The object is at a distance $r = R + 3R = 4R$ from the center of the Earth.
For the object to not return to Earth, its total mechanical energy must be at least zero at infinity, where potential and kinetic energy are both zero.
Applying the principle of conservation of energy between the starting point and infinity:
$K_i + U_i = K_f + U_f \implies \frac{1}{2}mv^2 - \frac{GMm}{4R} = 0 + 0$
Solving for velocity $v$:
$\frac{1}{2}mv^2 = \frac{GMm}{4R} \implies v^2 = \frac{2GM}{4R} = \frac{GM}{2R}$
Thus, the minimum escape speed is $v = \sqrt{\frac{GM}{2R}}$.

Angular momentum, $L=m V_0 R=m {\sqrt{G M R}}$ where $M$ is the mass of star$\;\frac{L_B}{L_A}=\;\frac{m_B}{m_A} \sqrt{\;\frac{R_B}{R_A}}=8$

The escape energy $K$ required to project a body from the earth's surface ($r=R$) to infinity is $K = U_{\infty } - U_{surface} = 0 - (-\frac{G M m}{R}) = \frac{G M m}{R}$. Using the relation $g = \frac{G M}{R^2}$, we substitute $G M = g R^2$ into the expression to get $K = \frac{(g R^2) m}{R} = m g R$. Since the assertion states $K = \frac{1}{2} m g R$, Assertion A is false. The potential energy of the body is $U(r) = -\frac{G M m}{r}$, which approaches $0$ as $r \to \infty$. Since $U(r)$ is negative for all finite values of $r$, $0$ is indeed the maximum value of potential energy. Thus, Reason R is true.

As limiting case of elliptical path is straight line therefore proportionality $T^2 \propto \;\frac{a^3}{M}$ will holds from$\; (T^2=\;\frac{4 \pi ^2}{G M} r^3)$ $\;\;\frac{4^2}{T^2}=\;\frac{a^3}{m}(2 a)^3=\;\frac{2}{8}=\;\frac{1}{4}$ $\;\;\frac{4}{T}=\;\frac{1}{2}$ $\;T=8 \;\text{ seconds }\;$

$\;T_1 \propto (R)^{3 / 2}$ $\;\;\text{ and }\; T_2 \propto (1.03 R)^{3 / 2}$ $\;\Rightarrow \;\; T_2=(1.03 R)^{3 / 2} \cdot T_1 \approx 1.045 T_1$So $T_2$ will larger by $4.5 \%$ w.r.t. $T_1$.

&g=\;\frac{G M}{R^2} $\;\text{ Now, }\; g^{'}=\;\frac{G M}{ (\;\frac{R}{3}) ^2}=\;\frac{9 G M}{R^2}=9 g$

$L=m v r$ $L\;=m \sqrt{\;\frac{G M}{R} R}$ $\;=m {\sqrt{G M R}}$ $\;=\;\frac{M}{2} \sqrt{G M \;\frac{4 R}{3}}$ $\;=M \sqrt{\;\frac{G M R}{3}}$ $x=3$ (But for comparable mass, this solution is not applicable)Alternate solution:$\;m_1 r_1=m_2 r_2 \Rightarrow r_1=\;\frac{8 R}{9}$ $\;\;\frac{G M \frac{M}{2}}{ (\;\frac{4 R}{3}) ^2}=\;\frac{M}{2} \omega ^2 \;\frac{8 R}{9}$ $\;\omega =\sqrt{\;\frac{81 G M}{128 R^3}}$ $\;L=I \omega =\;\frac{M}{2} (\;\frac{8 R}{9}) ^2 \omega$ $\;=M \sqrt{G M R \;\frac{8}{81}}$ $\;\Rightarrow x \approx 10$

$K\;=-\;\frac{U}{2}=\;\frac{G M m}{2 r}$ $\;=\;\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^3}{2 \times 6670 \times 10^3}$ $\;=3 \times 10^{10} {\text{J}}$

$\;F_1=\;\frac{G M m}{4 R^2}$ $\;F_2=\;\frac{G M m}{4 R^2}-G (\;\frac{M}{27}) \;\frac{m}{ (\;\frac{4 R}{3}) ^2}$ $\;=\;\frac{G M m}{R^2} (\;\frac{1}{4}-\;\frac{1}{48})$ $\;=\;\frac{11}{48} \;\frac{G M m}{R^2}$ $\;\;\frac{F_1}{F_2}=\;\frac{12}{11}$

$\;v_{\;\text{escape }\;}=\sqrt{\;\frac{2 G M}{R}} ; M_{\;\text{earth }\;}=8 M_{\;\text{planet }\;} ; R_{\;\text{earth }\;}=2 R_{\;\text{planet }\;}$ $\;\;{v_{\;\text{planet }\;}}/{v_{\;\text{earth }\;}}=\sqrt{\;{M_{\;\text{planet }\;}}/{M_{\;\text{earth }\;}} \times \;{R_{\;\text{earth }\;}}/{R_{\;\text{planet }\;}}}$ $\;v_{\;\text{planet }\;}=(11.2 km / s ) \sqrt{\;\frac{1}{8} \times 2}$ $\;=5.6 km / s$

$\;F_{\;\text{net }\;}={\sqrt{2}} F+F^{'}$ $\;F=\;\frac{G m^2}{a^2} \;\text{ and }\; F^{'}=\;{G m^2}/{({\sqrt{2}} {\text{a}})^2}$ $\;F_{\;\text{net }\;}={\sqrt{2}} \;{ {\text{Gm}}^2}/{ {\text{a}}^2}+\;{ {\text{Gm}}^2}/{2 {\text{a}}^2}$ $\; (\;{2 {\sqrt{2}}+1}/{32}) \;{ {\text{Gm}}^2}/{ {\text{L}}^2}=\;{ {\text{Gm}}^2}/{ {\text{a}}^2} (\;{2 {\sqrt{2}}+1}/{2})$ $\; {\text{a}}=4 {\text{L}}$

$\; {\text{T}}^2 \propto {\text{r}}^3$ $\;\;{ {\text{T}}_1^2}/{ {\text{r}}_1^3}=\;{ {\text{T}}_2^2}/{ {\text{r}}_2^3}$ $\;\;{(200)^2}/{ {\text{r}}^3}=\;{ {\text{T}}_2^2}/{ (\;{ {\text{r}}}/{4}) ^3}$ $\;\;\frac{200 \times 200}{4 \times 4 \times 4}= {\text{T}}_2^2$ $\; {\text{T}}_2=\;\frac{200}{4 \times 2}$ $\; {\text{T}}_2=25 \;\text{ days }\;$