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.

Revise with Concepts

Escape Speed

Earth Satellites

Orbital Velocity

Geostationary Satellite - Medium

Energy of an Orbiting Satellite

Geostationary and Polar Satellites

Quick Summary With Stories

Escape Velocity and its Magnitude

Introduction to Satellites.

Types of Satellite

Introduction To Time Period and Velocity of Satellite

Orbital Velocity

Condition for Trajectories of Projected Bodies

Geostationary Satellite

Conservation of Angular Momentum for Masses in Orbit

Polar Satellites

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?

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

What is escape velocity? Obtain an expression for it.

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:

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

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:

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

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?

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

Revise with Concepts

Escape Speed

Earth Satellites

Orbital Velocity

Geostationary Satellite - Medium

Energy of an Orbiting Satellite

Geostationary and Polar Satellites

Quick Summary With Stories

Escape Velocity and its Magnitude

Introduction to Satellites.

Types of Satellite

Introduction To Time Period and Velocity of Satellite

Orbital Velocity

Condition for Trajectories of Projected Bodies

Geostationary Satellite

Conservation of Angular Momentum for Masses in Orbit

Polar Satellites

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?

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:

What is escape velocity? Obtain an expression for it.

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

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

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:

An artificial satellite moving in circular orbit around the earth has a total ( kinetic + potential ) energy E0​. Its potential energy and kinetic energy respectively are

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?

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 g0​, value of acceleration due to gravity at the earth's surface, is

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 =6.0×1024 kg; radius of the earth =6.4×106m; G = 6.67×10−11 Nm2kg−2.

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

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:

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

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

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?

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