All Solutions
Section 9.4: Planetary Motion and Orbits
According to Kepler’s first law:
Planets follow elliptical orbits, with the Sun at one focus of the ellipse.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
According to Kepler’s second law which assigns that as a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time.
Solve for first case (before the radius increase):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{1} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
end{align*}
$$
Solve for second case (after the radius increase):
Let the factor of increasing is $k$. As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{2} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{2}^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( k ~ r_{1} right)^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( k right)^{dfrac{3}{2}} ~ r_{1}^{dfrac{3}{2}} \
&=left( k right)^{dfrac{3}{2}} ~ left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
&= left( k right)^{dfrac{3}{2}} ~ T_{1} \
end{align*}
$$
So, if the radius of the orbital is increasing, then the period of the orbital increases by the factor $left( k right)^{dfrac{3}{2}}$.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
So, the orbital period of the planet doesn’t depend on the planet’s mass itself. But the orbital period of the planet depends on the mass of the affecting star and the radius of the orbital.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the star and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{M} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
As we know the geosynchronous satellite means that the satellite takes $24 mathrm{~hr}$ to complete one cycle.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the star and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{M} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
As we know the geosynchronous satellite means that the satellite takes $24 mathrm{~hr}$ to complete one cycle. So, the satellite must orbit about a planet, not on a part of the planet. So, the answer is “No, it isn’t possible to put a satellite into an orbit such that it remains fixed above the North Pole. “
Solve for first case (before the radius quadrupled):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{1} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
end{align*}
$$
Solve for second case (after the radius quadrupled):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{2} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{2}^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( 4 r_{1} right)^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( 4 right)^{dfrac{3}{2}} ~ r_{1}^{dfrac{3}{2}} \
&= 8 ~ left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
&= 8 ~ T_{1} \
end{align*}
$$
So, The answer is “Yes, the period of the satellite increases.”
Solve for first case (before the radius quadrupled):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{1} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
end{align*}
$$
Solve for second case (after the radius quadrupled):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T_{2} &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{2}^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( 4 r_{1} right)^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ left( 4 right)^{dfrac{3}{2}} ~ r_{1}^{dfrac{3}{2}} \
&= 8 ~ left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r_{1}^{dfrac{3}{2}} \
&= 8 ~ T_{1} \
end{align*}
$$
So, if the radius of the satellite is quadrupled, then the period will increase by factor of $8$.
Solve for the first case (before the radius quadrupled):
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
Similarly, the speed of the satellite that orbiting Earth is given by
$$
begin{align*}
v_{1} &= sqrt{ dfrac{ G ~ M_{E} }{ r_{1} } }\
end{align*}
$$
Solve for the second case (after the radius quadrupled):
Also, the speed of the satellite that orbiting Earth is given by
$$
begin{align*}
v_{2} &= sqrt{ dfrac{ G ~ M_{E} }{ r_{2} } }\
&= sqrt{ dfrac{ G ~ M_{E} }{ 4 ~ r_{1} } } \
&= dfrac{1}{2} ~ sqrt{ dfrac{ G ~ M_{E} }{ r_{1} } } \
&= dfrac{1}{2} ~ v_{1}
end{align*}
$$
So, if the radius of the satellite is quadrupled, then the speed of the satellite will decrease by factor of $2$.
(b) If the radius of the satellite is quadrupled, then the period will increase by a factor of $8$.
(c) If the radius of the satellite is quadrupled, then the speed of the satellite will decrease by a factor of $2$.
The period that Venus takes to rotate about the sun is $T = 1.93 times 10^{7} mathrm{~s}$. The mass of the Sun is $M_{s} = 1.989 times 10^{30} mathrm{~kg}$.
$textbf{Required: }$
Finding the average distance from the Sun.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
Rearrange and solve for the average distance from the Sun:
$$
begin{align*}
r^{dfrac{3}{2}} &= dfrac{ T }{ left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) } \
r &= left( dfrac{ T }{ left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) } right)^{dfrac{2}{3}} \
&= left( dfrac{ T ~ sqrt{ G ~ M_{S} } }{ 2 pi } right)^{dfrac{2}{3}} \
&= left( dfrac{ 1.93 times 10^{7} mathrm{~s} times sqrt{ 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} times 1.989 times 10^{30} mathrm{~kg} } }{ 2 pi } right)^{dfrac{2}{3}} \
&= 1.078 times 10^{11} mathrm{~m}
end{align*}
$$
So, the average distance from the Sun is $1.078 times 10^{11} mathrm{~m}$.
The mass of the Moon is $M_{M} = 7.348 times 10^{22} mathrm{~kg}$. The radius of the Moon is $r_{M} = 1.737 times 10^{6} mathrm{~m}$. The height of the spacecraft is $h = 1.1 times 10^{5} mathrm{~m}$.
$textbf{Required: }$
Finding the period of completing one cycle that it take for the command module to complete one orbit.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Moon and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{M} } } right) ~ r^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{M} } } right) ~ left( r_{M} + h right)^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} times 7.348 times 10^{22} mathrm{~kg}} } right) ~ left( 1.737 times 10^{6} mathrm{~m} + 1.1 times 10^{5} mathrm{~m} right)^{dfrac{3}{2}} \
&= 7.124 times 10^{3} mathrm{~s}
end{align*}
$$
So, the period of completing one cycle that it takes for the command module to complete one orbit is $7.124 times 10^{3} mathrm{~s}$.
The mass of the Sun is $M_{s} = 1.989 times 10^{30} mathrm{~kg}$. The average distance between Jupiter and the Sun is $r = 760.15 mathrm{~million ~ km}$.
$textbf{Required: }$
Finding the period of completing one cycle that it takes for the Jupiter around the Sun.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
&= left( dfrac{ 2 pi }{ sqrt{ 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} times 1.989 times 10^{30} mathrm{~kg}} } right) ~ left( 7.602 times 10^{11} mathrm{~m} right)^{dfrac{3}{2}} \
&= 3.616 times 10^{8} mathrm{~s}
end{align*}
$$
So, the period of completing one cycle that it takes for the Jupiter around the Sun is $3.616 times 10^{8} mathrm{~s}$.
Earth’s radius is $R_{E} = 6.37 times 10^{6} mathrm{~m}$. The height of the satellite above the Earth is $h = 2.02 times 10^{7} mathrm{~m}$. The mass of Earth is $m_{E} = 5.972 times 10^{24} mathrm{~kg}$.
$textbf{Required: }$
Finding the speed of the satellite.
As the textbook mentions Kepler’s third law that studied the relation between the distance of a planet from the Sun and its orbital period which is given by
$$
begin{align*}
T &= left( dfrac{ 2 pi }{ sqrt{ G ~ M_{S} } } right) ~ r^{dfrac{3}{2}} \
end{align*}
$$
Similarly, the speed of the satellite that orbiting Earth is given by
$$
begin{align*}
v &= sqrt{ dfrac{ G ~ M_{E} }{ r } }\
&= sqrt{ dfrac{ G ~ M_{E} }{ R_{E} + h } } \
&= sqrt{ dfrac{ 6.67 times 10^{-11} mathrm{~N cdot m^{2}/kg^{2}} times 5.972 times 10^{24} mathrm{~kg} }{ 6.37 times 10^{6} mathrm{~m} + 2.02 times 10^{7} mathrm{~m} } }\
&= 3.872 times 10^{3} mathrm{~m/s}
end{align*}
$$
So, the speed of the satellite is $3.872 times 10^{3} mathrm{~m/s}$.
