All Solutions
Page 174: Practice Problems
{color{#c34632}bf{}This is solution is for you If the period is color{#4257b2}7.15 days }
$$
$$
text{color{#4257b2}According to Kepler’s 3rd law $$left(dfrac{T_G}{T_I}right)^2=left(dfrac{r_G}{r_I}right)^3$$}
$$
We need to solve for $r_G$
Raise the power on both sides to $1/3$
$$
left(dfrac{T_G}{T_I}right)^{2/3}=left(dfrac{r_G}{r_I}right)
$$
$$
r_G = r_Ileft(dfrac{T_G}{T_I}right)^{2/3}
$$
From example 1 we can use: $r_1 = 4.2, T_I=1.8$, it is given that $T_G = 7.15$
$$
r_G = 4.2cdot left(dfrac{7.15}{1.8}right)^{2/3}approx 10.5
$$
{color{#c34632}bf{}This is solution is for you If the period is color{#4257b2}32 days }
$$
$$
text{color{#4257b2}According to Kepler’s 3rd law $$left(dfrac{T_G}{T_I}right)^2=left(dfrac{r_G}{r_I}right)^3$$}
$$
We need to solve for $r_G$
Raise the power on both sides to $1/3$
$$
left(dfrac{T_G}{T_I}right)^{2/3}=left(dfrac{r_G}{r_I}right)
$$
$$
r_G = r_Ileft(dfrac{T_G}{T_I}right)^{2/3}
$$
From example 1 we can use: $r_1 = 4.2, T_I=1.8$, it is given that $T_G = 32$
$$
r_G = 4.2cdot left(dfrac{32}{1.8}right)^{2/3}approx 28.6
$$
text{color{#4257b2}bf{}SEE SOLUTION}
$$
**Given values.**
$T_c=7.15 , text{days}$
$T_l=1.8 , text{days}$
$r_l=4.2 , text{units}$
***
**Kepler’s third law.**
–
$$bigg(dfrac{T_c}{T_l} bigg)^2=bigg(dfrac{r_c}{r_l} bigg)^3. tag1$$
When we rearrange equation (1), we get.
$$r_c^3=r_{l}^3 cdot bigg(dfrac{T_c}{T_l} bigg)^2 tag2$$
By further rearranging, we get.
$$r_c=sqrt[3]{r_{l}^3 cdot bigg(dfrac{T_c}{T_l} bigg)^2} tag3$$
***
$$begin{aligned}
r_c&=sqrt[3]{(4.2 , text{units})^3 bigg(dfrac{7.15 , text{days}}{1.8 , text{days}} bigg)^2}\
r_c&=sqrt[3]{(74.088 , text{units}^3)(15.78)}\
r_c&=sqrt[3]{1169.01 , text{units}^3}\
r_c &=color{#4257b2}{10.53 , text{units}}
end{aligned}$$
**Given values.**
$T_c=32 , text{days}$
$T_l=1.8 , text{days}$
$r_l=4.2 , text{units}$
***
**Kepler’s third law.**
–
$$bigg(dfrac{T_c}{T_l} bigg)^2=bigg(dfrac{r_c}{r_l} bigg)^3. tag1$$
When we rearrange equation (1), we get.
$$r_c^3=r_{l}^3 cdot bigg(dfrac{T_c}{T_l} bigg)^2 tag2$$
By further rearranging, we get.
$$r_c=sqrt[3]{r_{l}^3 cdot bigg(dfrac{T_c}{T_l} bigg)^2} tag3$$
***
$$begin{aligned}
r_c&=sqrt[3]{(4.2 , text{units})^3 bigg(dfrac{32 , text{days}}{1.8 , text{days}} bigg)^2}\
r_c&=sqrt[3]{(74.088 , text{units}^3)(316.04)}\
r_c&=sqrt[3]{23415.5 , text{units}^3}\
r_c &=color{#c34632}{28.6 , text{units}}
end{aligned}$$
$$
begin{aligned}
left( frac{T_{a}}{T_{E}} right)^{2}=left( frac{r_{a}}{r_{E}}right)^{3}
end{aligned}
$$
where $T_{a}$ is a period of an asteriod, $T_{E}$ is a period of Earth, $r_{a}$ is a radius of an asteroid, and $r_{E}$ is a radius of Earth.
$$
begin{aligned}
r_{a}=2r_{E}
end{aligned}
$$
$$
begin{aligned}
left( frac{T_{a}}{T_{E}} right)^{2}&=left( frac{2r_{E}}{r_{E}}right)^{3}\
left( frac{T_{a}}{T_{E}} right)^{2}&= 2^{3}\
left( frac{T_{a}}{T_{E}} right)^{2}&= 8\
end{aligned}
$$
$$
begin{aligned}
left( frac{T_{a}}{T_{E}} right)^{2}&= 8 \
sqrt{left( frac{T_{a}}{T_{E}} right)^{2}}&= sqrt{8}\
frac{T_{a}}{T_{E}} &=2.83
end{aligned}
$$
$$
begin{aligned}
frac{T_{a}}{T_{E}} &=2.83 / cdot T_{E}\
T_{a}&=2.83T_{E}\
T_{a}&=2.83cdot 1 hspace{0.5mm} mathrm{yr}\
T_{a}&=2.83 hspace{0.5mm} mathrm{yr}\
end{aligned}
$$
$$
begin{align*}
left(frac{T_E}{T_M}right)^2&=left(frac{r_E}{r_M}right)^3
end{align*}
$$
where the $r_M$ stands for the radius of the Mars orbit, $r_E$ represents the radius of the Earth radius, also $T_E$ and $T_M$ stands for the time needed for the Earth and Mars to orbit over the earth.
We know that the time needed for Earth to orbit over the Sun is $T_E=365text{ days}$. We have given the relation of radius of orbite $r_M=1.52cdot{r_E}$.
$$
begin{align*}
left(frac{T_E}{T_M}right)^2&=left(frac{r_E}{r_M}right)^3\
T_M&=sqrtfrac{(T_E)^2}{left(frac{r_E}{r_M}right)^3}\
T_M&=sqrtfrac{(365text{ days})^2}{left(frac{r_E}{1.52cdot{r_E}}right)^3}\
T_M&=sqrtfrac{(365text{ days})^2}{left(frac{1}{1.52}right)^3}
end{align*}
$$
$$
boxed{T_M=684text{ days}}
$$
T_M=684text{ days}
$$
$T_e=365 , text{days}$
$R_M=1.52 cdot R_E$
***
**Kepler’s third law.**
–
$$bigg(dfrac{T_e}{T_M} bigg)^2=bigg(dfrac{R_e}{R_M} bigg)^3, tag1$$
The squared quantity of the period of the earth divided by the period of the Moon is equal to the cubed quantity of the earth’s average distance from the Sun, divided by the Moon’s average distance from the Sun.
$$T_M^2=T_e^2 cdot bigg(dfrac{R_M}{R_E} bigg)^3 tag2$$
Since the $R_M=1.52 cdot R_E$, equation (2) becomes.
$$T_M^2=T_e^2 cdot bigg(dfrac{1.52 cdot R_E}{R_E} bigg)^3 tag3$$
By further rearranging we get.
$$ T_M=sqrt{T_e^2 cdot bigg(1.52 bigg)^3} tag4$$
Plug in values into equation (4) and calculate.
***
$$begin{aligned}
T_{M} &=sqrt{(365 , text{days})^2 cdot (1.52)^3 }\
T_{M} &=sqrt{467860.6 , text{days}^2 }\
T_{M} &=color{#c34632}{684 , text{days} }
end{aligned}$$
$T_M=27.3 , text{days}$
$R_E=6.38 cdot 10^{3} , text{km}$
$R_M=3.9 cdot 10^{5} , text{km}$
***
**Kepler’s third law.**
–
$$bigg(dfrac{T_s}{T_M} bigg)^2=bigg(dfrac{r_s}{R_M} bigg)^3, tag1$$
The squared quantity of the period of the satellite divided by the period of the Moon is equal to the cubed quantity of the satellite’s average distance from the Sun, divided by the Moon’s average distance from the Sun.
$$T_s^2=T_M^2 bigg(dfrac{r_s}{R_M} bigg)^3 tag2$$
Plug in values into equation (2) and calculate.
**Calculation method.**
$$begin{aligned}
T_{s} &=sqrt{(27.3 , text{days})^2 bigg(dfrac{6.7 cdot 10^{3} , text{km}}{3.9 cdot 10^{5} , text{km}} bigg)^3}\
T_{s} &=sqrt{(745.29 , text{days}^2)(5.07 cdot 10^{-6})}\
T_{s} &=sqrt{0.00377 , text{days}^2}\
T_{s} &=color{#c34632}{0.06147 , text{day}}
end{aligned}$$
$$h=r_s-r_E tag3$$
Plug in values into equation (3) and calculate.
***
$$begin{aligned}
h &=(6.7 cdot 10^{3} , text{km})-(6.38 cdot 10^{3} , text{km})\
h &=color{#c34632}{320 , text{km}}
end{aligned}$$
b) $320 , text{km}$
Solve Kepler’s third law for the period of the satelite, $T_s$:
$T_s^2 = T_M^2 (dfrac{r_s}{r_M})^3$
$T_s^2 = (27.3)^2 (dfrac{6.70e3}{3.90e5})^3$
$T_s^2 = 0.0037788$
Thus:
$T_s = 0.0615 day$
b)
The radius of Earth is $6.37times 10^3$ km, thus the height of the satellite is:
$$
h = (6.70times 10^3) – (6.37times 10^3) = 3.3 times 10^2 km
$$
b) $3.3 times 10^2 km$
$$
begin{align}
left(frac{T_M}{T_S}right)^2&=left(frac{r_M}{r_S}right)^3
end{align}
$$
Here the satellite and the moon orbits around the Earth. $T_M$ and $T_S$ stand for the time needed for the moon and satellite to make one complete rotation around the Earth. The $r_M$ and $r_S$ represent the radius of orbit for the Moon and satellite around the earth. We have given:
$$
begin{align*}
T_M&=27.3text{ days}\
T_S&=1text{ day}\
r_M&=3.9cdot{10^5}text{ km}
end{align*}
$$
Let’s express $r_S$ from the first equation and substitute:
$$
begin{align*}
r^3_S&=frac{r^3_M}{left(frac{T_M}{T_S}right)^2}\
r^3_S&=frac{T^2_Scdot{r^3_M}}{T^2_M}\
r_S&=sqrt[3]{frac{T^2_Scdot{r^3_M}}{T^2_M}}\
r_S&=sqrt[3]{frac{(1text{ day})^2cdot{(3.9cdot{10^5}text{ km})^3}}{(27.3text{ days})^2}}
end{align*}
$$
$$
boxed{r_S=43000text{ km}}
$$
r_S=43000text{ km}
$$
$T_M=27.3 , text{days}$
$T_s=1 , text{day}$
$R_M=3.9 cdot 10^{5} , text{km}$
***
**Kepler’s third law.**
–
$$bigg(dfrac{T_M}{T_s} bigg)^2=bigg(dfrac{R_m}{R_s} bigg)^3, tag1$$
The squared quantity of the period of the moon divided by the period of the satellite is equal to the cubed quantity of the Moon’s average distance from the Sun, divided by the satellite’s average distance from the Sun.
$$R_s^3=dfrac{R_M^3}{bigg(dfrac{T_M}{T_S} bigg)^2}tag2$$
By further derivation, we get.
$$R_s^{3}=dfrac{T_{s}^2 cdot R_{M}^3}{T_{M}^2} tag3$$
Final relation will be.
$$R_s =sqrt[3]{dfrac{T_{s}^2 cdot R_{M}^3}{T_{M}^2}} tag4$$
***
**Calculation method.**
$$begin{aligned}
R_s &=sqrt[3]{dfrac{ (1 , text{day})^2 cdot (3.9 cdot 10^{5} , text{km})^3}{(27.3 , text{days})^2}}\
R_s &=sqrt[3]{dfrac{(5.93 cdot 10^{16} , text{km}^3 cdot cancel{text{day}^2})}{745.29 , cancel{text{days}^2}}}\
R_s &=sqrt[3]{7.95 cdot 10^{13} , text{km}^3}\
R_{s} &=color{#c34632}{43 010 , text{km}}
end{aligned}$$
