Q: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Given that, Mass of satellite \$=m\$ Mass of planet \$=M\$ Radius \$=R\$ Altitude \$h=2R\$ Now, The gravitational potential energy \$P.E=\dfrac{-Gm}{r}\$ Potential energy at altitude \$=\dfrac{GmM}{3R}\$ Orbital velocity \${{v}_{0}}=\dfrac{GmM}{R+h}\$ Now, the total energy is \$ {{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R} \$ \$ {{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R} \$ \$ {{E}_{f}}=\dfrac{GmM}{3R}\left[ \dfrac{1}{2}-1 \right] \$ \$ {{E}_{f}}=\dfrac{-GmM}{6R} \$ Now, \${{E}_{i}}={{E}_{f}}\$ Now, the minimum required energy \$ K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R} \$ \$ K.E=\dfrac{5GmM}{6R} \$ Hence, the minimum required energy is \$\dfrac{5GmM}{6R}\$

Q: A geostationary satellite is orbiting the earth at a height \$6R\$ above the earth's surface, where \$R\$ is radius of earth. The time period of another satellite at a height \$2.5R\$ from earth's surface would be:

For geostationary satellite : \$r_1 = R + 6 R = 7R\$ \$T_1 = 24\$ \$h\$For second satellite : \$r_2 = R + 2.5 R = 3.5R\$ Time period of satellite : \$T = 2\pi \sqrt{\dfrac{r^3}{GM}}\$ where, \$r\$ is the radius of orbit\$\implies\$ \$\dfrac{T_2}{T_1} = \sqrt{\dfrac{r^3_2}{r^3_1}}\$\$\therefore\$ \$\dfrac{T_2}{24} = \sqrt{\dfrac{(3.5R)^3}{(7R)^3}} = \dfrac{1}{2 \sqrt{2}}\$\$\implies\$ \$T_2 = \dfrac{24}{2\sqrt{2}} = 6\sqrt{2}\$ \$h\$

Q: The ratio of escape velocity at earth\$(v_e)\$ to the escape velocity at a planet\$(v_p)\$ whose radius and mean density are twice as that of earth is:

We know that: \$v_e=\sqrt{\dfrac{2GM}{R}}\$Given that: \$R'=2R\$ and \$\rho'=2\rho\$\$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho\$\$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M\$\$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}\$\$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}\$

Q: Two satellites \$A\$ and \$B\$ go round the planet \$P\$ in circular orbits having radii \$4R\$ and \$R\$ respectively. If the speed of the satellite \$A\$ is \$3v\$, then the speed of satellite \$B\$ will be

Velocity \$v\$ of a satellite varies inversely as the square root of the orbit of radius \$r\$\$\quad v\propto \cfrac { 1 }{ \sqrt { r } } \$\$\therefore \cfrac { { V }_{ A } }{ { V }_{ B } } =\sqrt { \cfrac { { r }_{ B } }{ { r }_{ A } } } =\sqrt { \cfrac { R }{ 4R } } \$\$\Rightarrow \cfrac { 3v }{ { V }_{ B } } =\cfrac { 1 }{ 2 } \Rightarrow { V }_{ B }=6v\$

Q: The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. if the speed of satellite A is 3v, then the speed of satellite B will be:

\$=\sqrt { \cfrac { GM }{ R } } \\ \cfrac { { V }_{ A } }{ { V }_{ B } } =\sqrt { \cfrac { { r }_{ B } }{ { r }_{ A } } } =\sqrt { \cfrac { R }{ 4R } } =\cfrac { 1 }{ 2 } \\ \therefore \cfrac { { V }_{ A } }{ { V }_{ B } } =\cfrac { 3V }{ { V }_{ B } } =\cfrac { 1 }{ 2 } \\ \therefore { V }_{ B }=6V\$

Q: The period of revolution of planet A around the sun is \$8\$ times that of B. The distance of A from the sun is how many times greater than that of B from the sun?

Given,\$T_A=8T_B\$According to Kepler's third law,\$(\dfrac{T_A}{T_B})^2=(\dfrac{R_A}{R_B})^3\$\$\dfrac{R_A}{R_B}=(\dfrac{T_A}{T_B})^{\dfrac 23}\$\$\dfrac{R_A}{R_B}=(\dfrac{8T_B}{T_B})^{\dfrac 23}\$\$\dfrac{R_A}{R_B}=2^{3\times \dfrac 23}=2^2=4\$\$R_A=4R_B\$So the correct answer is \$4\$.

Q: A stellite of mass m is orbiting the earth (of radius \$R\$) at a height \$h\$ from its surface. The total energy of the stellite in terms of \$g_0\$, value of acceleration due to gravity at the earth's surface, is

Total energy of the satellite \$T = \dfrac{-GMm}{2r}\$ where \$r = R+h\$\$\therefore\$ \$T = \dfrac{-GMm}{2 (R+h)}\$Acceleration due to gravity at earth's surface \$g_o = \dfrac{GM}{R^2}\$ \$\implies GM = g_oR^2\$\$\implies\$ \$T = \dfrac{-mg_o R^2}{2 (R+h)}\$

Q: A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \$= 200\ kg\$; mass of the earth \$= 6.0 \times 10^{24}\ kg\$; radius of the earth \$=6.4 \times 10^6\, m\$; G = \$6.67 \times 10^{-11}\ Nm^2kg^{-2}\$.

Mass of the Earth, \$M=6.0 \times 10^{24}\ kg\$\$m=200\ kg\$ \$R_e=6.4 \times 10^6\ m\$\$G=6.67 \times 10^{-11}\ Nm^2kg^{-2}\$Height of the satellite,\$h=400\ km = 4 \times 10^5\ m\$Total energy of the satellite at height h\$ = (1/2)mv^2+\left(-G \dfrac{M_em}{(R_e+h)}\right)\$Orbital velocity of the satellite, \$v=\sqrt{\dfrac{GM_e}{R_e+h}}\$Total energy at height h \$=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}\$\$Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}\$The negative sign indicates that the satellite is bound to the Earth. Energy required to send the satellite out of its orbit = – (Bound energy)\$=\dfrac{GM_em}{2(R_e+h)}\$\$=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}\$\$=5.9 \times 10^9\ J\$If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply \$5.9 \times 10^9J\$ of energy to just escape it.

Question 1

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Accepted Answer

Given that,&#10;&#10;Mass of satellite $=m$ &#10;&#10;Mass of planet $=M$ &#10;&#10;Radius $=R$ &#10;&#10;Altitude $h=2R$ &#10;&#10;Now,&#10;&#10;The gravitational potential energy &#10;&#10;$P.E=\dfrac{-Gm}{r}$ &#10;&#10;Potential energy at altitude $=\dfrac{GmM}{3R}$ &#10;&#10;Orbital velocity ${{v}_{0}}=\dfrac{GmM}{R+h}$ &#10;&#10;Now, the total energy is &#10;&#10;  $ {{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R}&#10;$ &#10;&#10; $ {{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R}&#10;$ &#10;&#10; $ {{E}_{f}}=\dfrac{GmM}{3R}\left[&#10;\dfrac{1}{2}-1 \right] $ &#10;&#10; $ {{E}_{f}}=\dfrac{-GmM}{6R}&#10;$ &#10;&#10;Now, ${{E}_{i}}={{E}_{f}}$&#10;&#10;Now, the minimum required energy &#10;&#10;  $ K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R}&#10;$ &#10;&#10; $ K.E=\dfrac{5GmM}{6R}&#10;$ &#10;&#10;Hence, the minimum required energy is $\dfrac{5GmM}{6R}$

Question 2

A geostationary satellite is orbiting the earth at a height $6R$ above the earth's surface, where $R$ is radius of earth. The time period of another satellite at a height $2.5R$ from earth's surface would be:

Accepted Answer

For geostationary satellite :        $r_1  = R + 6 R   = 7R$            $T_1 = 24$ $h$For second satellite :     $r_2  = R + 2.5 R   = 3.5R$            Time period of satellite :       $T = 2\pi \sqrt{\dfrac{r^3}{GM}}$          where,  $r$ is the radius of orbit$\implies$    $\dfrac{T_2}{T_1}   = \sqrt{\dfrac{r^3_2}{r^3_1}}$$\therefore$       $\dfrac{T_2}{24}   = \sqrt{\dfrac{(3.5R)^3}{(7R)^3}}   = \dfrac{1}{2 \sqrt{2}}$$\implies$   $T_2   = \dfrac{24}{2\sqrt{2}}   = 6\sqrt{2}$ $h$

Question 3

What is escape velocity? Obtain an expression for it.

Accepted Answer

Escape velocity: -The minimum speed with which a body must be projected in order that it will escape from the earth gravitational filed is called as escape velocity.Expression of it:-consider a body of mass $m$ on the surface of earthwhose mass and radius are respectively represented by $M$ and $R$.and earth is $F = \dfrac{{GMm}}{{{R^2}}}$gravity potential energy $\left( {GPE} \right) = \dfrac{{GMm}}{{{R^2}}}$$w = F.S = \dfrac{{GMm}}{{{R^2}}}$In order to projector body into the space overcoming the earth's gravitational field it must be give a kinetic energy which is equal topotential energy.therefore;kinetic energy is equal to potential energy.

Question 4

The ratio of escape velocity at earth$(v_e)$ to the escape velocity at a planet$(v_p)$ whose radius and mean density are twice as that of earth is:

Accepted Answer

We know that: $v_e=\sqrt{\dfrac{2GM}{R}}$Given that: $R'=2R$ and $\rho'=2\rho$$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho$$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M$$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}$$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}$

Question 5

Two satellites $A$ and $B$ go round the planet $P$ in circular orbits having radii $4R$ and $R$ respectively. If the speed of the satellite $A$ is $3v$, then the speed of satellite $B$ will be

Accepted Answer

Velocity $v$ of a satellite varies inversely as the square root of the orbit of radius $r$$\quad v\propto \cfrac { 1 }{ \sqrt { r }  } $$\therefore \cfrac { { V }_{ A } }{ { V }_{ B } } =\sqrt { \cfrac { { r }_{ B } }{ { r }_{ A } }  } =\sqrt { \cfrac { R }{ 4R }  } $$\Rightarrow \cfrac { 3v }{ { V }_{ B } } =\cfrac { 1 }{ 2 } \Rightarrow { V }_{ B }=6v$

Question 6

The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. if the speed of satellite A is 3v, then the speed of satellite B will be:

Accepted Answer

$=\sqrt { \cfrac { GM }{ R }  } \ \cfrac { { V }_{ A } }{ { V }_{ B } } =\sqrt { \cfrac { { r }_{ B } }{ { r }_{ A } }  } =\sqrt { \cfrac { R }{ 4R }  } =\cfrac { 1 }{ 2 } \ \therefore \cfrac { { V }_{ A } }{ { V }_{ B } } =\cfrac { 3V }{ { V }_{ B } } =\cfrac { 1 }{ 2 } \ \therefore { V }_{ B }=6V$

Question 7

An artificial satellite moving in circular orbit around the earth has a total $($ kinetic + potential $)$ energy $E_{0}$. Its potential energy and kinetic energy respectively are

Accepted Answer

From the options itself it is clear that only in option C, when you add PE and KE i.e $2E_o+(-E_o)$ we get $E_o$NowTotal energy $E_o=P.E+K.E=\dfrac {-GMm}{R}+\dfrac{1}{2}mv^2$When we substitute $v=\sqrt {\dfrac {GM}{R}}$ in K.E, we have $K.E=\dfrac {GMm}{2R}$$\therefore \dfrac {-GMm}{R}+\dfrac {GMm}{2R}=E_o$$\Rightarrow \dfrac {-GMm}{R}=2E_o$So, $P.E=\dfrac {-GMm}{R}=2E_o $ and  $K.E=\dfrac {GMm}{2R}=-E_o$

Question 8

The period of revolution of planet A around the sun is $8$ times that of B. The distance of A from the sun is how many times greater than that of B from the sun?

Accepted Answer

Given,$T_A=8T_B$According to Kepler's third law,$(\dfrac{T_A}{T_B})^2=(\dfrac{R_A}{R_B})^3$$\dfrac{R_A}{R_B}=(\dfrac{T_A}{T_B})^{\dfrac 23}$$\dfrac{R_A}{R_B}=(\dfrac{8T_B}{T_B})^{\dfrac 23}$$\dfrac{R_A}{R_B}=2^{3\times \dfrac 23}=2^2=4$$R_A=4R_B$So the correct answer is $4$.

Question 9

A stellite of mass m is orbiting the earth (of radius $R$ ) at a height $h$ from its surface. The total energy of the stellite in terms of $g_{0}$ , value of acceleration due to gravity at the earth's surface, is

Accepted Answer

Total energy of the satellite        $T  = \dfrac{-GMm}{2r}$             where  $r = R+h$$\therefore$     $T = \dfrac{-GMm}{2 (R+h)}$Acceleration due to gravity at earth's surface        $g_o = \dfrac{GM}{R^2}$         $\implies GM = g_oR^2$$\implies$     $T = \dfrac{-mg_o R^2}{2 (R+h)}$

Question 10

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite $= 200\ kg$; mass of the earth $= 6.0 \times 10^{24}\ kg$; radius of the earth $=6.4 \times 10^6\, m$; G = $6.67 \times 10^{-11}\ Nm^2kg^{-2}$.

Accepted Answer

Mass of the Earth, $M=6.0 \times 10^{24}\ kg$$m=200\ kg$ $R_e=6.4 \times 10^6\ m$$G=6.67 \times 10^{-11}\ Nm^2kg^{-2}$Height of the satellite,$h=400\ km = 4 \times 10^5\ m$Total energy of the satellite at height h$ = (1/2)mv^2+\left(-G \dfrac{M_em}{(R_e+h)}\right)$Orbital velocity of the satellite, $v=\sqrt{\dfrac{GM_e}{R_e+h}}$Total energy at height h $=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}$$Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}$The negative sign indicates that the satellite is bound to the Earth. Energy required to send the satellite out of its orbit = &#8211; (Bound energy)$=\dfrac{GM_em}{2(R_e+h)}$$=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}$$=5.9 \times 10^9\ J$If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply $5.9 \times 10^9J$ of energy to just escape it.

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