Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Textbook solutions
All Solutions
Page 195: Standardized Test Practice
Exercise 1
Step 1
1 of 4
In this problem, we need to find the period of the satellite. To do this, we are going to use Kepler’s third law, and we can write it as
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}=left( frac{r_{s2}}{r_{s1}}right)^{3}
end{aligned}
$$
where $T_{s1}$ is a period of the first satellite, $T_{s2}$ is a period of the second satellite, $r_{s1}$ is a radius of the first satellite’s orbit, and $r_{s2}$ is a radius of the second satellite’s orbit.
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}=left( frac{r_{s2}}{r_{s1}}right)^{3}
end{aligned}
$$
where $T_{s1}$ is a period of the first satellite, $T_{s2}$ is a period of the second satellite, $r_{s1}$ is a radius of the first satellite’s orbit, and $r_{s2}$ is a radius of the second satellite’s orbit.
Step 2
2 of 4
We know that the period of the first satellite is $T_{1}=1.0 cdot 10^{6} hspace{0.5mm} mathrm{s}$, and its radius of orbit is $r_{s1}=8.0 cdot 10^{6} hspace{0.5mm} mathrm{m}$. The orbital radius for the second satellite is $r_{s2}=2.0 cdot 10^{7} hspace{0.5mm} mathrm{m}$.
Step 3
3 of 4
Now, we can find the period for the second satellite
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}&=left( frac{r_{s2}}{r_{s1}}right)^{3}\
left( frac{T_{s2}}{T_{s1}} right)^{2}&=left( frac{2.0 cdot 10^{7} hspace{0.5mm} mathrm{m}}{8.0 cdot 10^{6} hspace{0.5mm} mathrm{m}}right)^{3}\
left( frac{T_{s2}}{T_{s1}} right)^{2}&=15.625\
T_{s2}&=sqrt{15.625}T_{s1}\
T_{s2}&=3.95cdot 10^{6} hspace{0.5mm} mathrm{s}
end{aligned}
$$
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}&=left( frac{r_{s2}}{r_{s1}}right)^{3}\
left( frac{T_{s2}}{T_{s1}} right)^{2}&=left( frac{2.0 cdot 10^{7} hspace{0.5mm} mathrm{m}}{8.0 cdot 10^{6} hspace{0.5mm} mathrm{m}}right)^{3}\
left( frac{T_{s2}}{T_{s1}} right)^{2}&=15.625\
T_{s2}&=sqrt{15.625}T_{s1}\
T_{s2}&=3.95cdot 10^{6} hspace{0.5mm} mathrm{s}
end{aligned}
$$
Result
4 of 4
$c)$ $4.0cdot 10^{6} hspace{0.5mm} mathrm{s}$
Exercise 2
Step 1
1 of 5
In this problem, we need to find the mass of the planet. To do this, we are going to use Newton’s version of Kepler’s third law. This law is the combination of Universal Gravitation and Kepler’s Third Law. We can write this law as
$$
begin{aligned}
T=2 pi sqrt{frac{r^{3}}{G m_{p}}}
end{aligned}
$$
where $T$ is a period of a satellite, $r$ is the radius of orbit of a satellite, $G$ is the gravitational constant, and $m_{S}$ is the mass of the planet. Also, we are going to use the equation
$$
begin{aligned}
T=frac{2 pi r}{v}
end{aligned}
$$
$$
begin{aligned}
T=2 pi sqrt{frac{r^{3}}{G m_{p}}}
end{aligned}
$$
where $T$ is a period of a satellite, $r$ is the radius of orbit of a satellite, $G$ is the gravitational constant, and $m_{S}$ is the mass of the planet. Also, we are going to use the equation
$$
begin{aligned}
T=frac{2 pi r}{v}
end{aligned}
$$
Step 2
2 of 5
We know the orbital radius, $r=6.7 cdot 10^{7} hspace{0.5mm} mathrm{m}$, and the speed of the satellite is $v=2.0 cdot 10^{5} hspace{0.5mm} mathrm{m/s}$. Also, we know the gravitational constant is $G=6.7 cdot 10^{-11}hspace{0.5mm} mathrm{frac{m^{3}}{kg s^{2}}}$.
Step 3
3 of 5
Now, we are going to combine these equations above
$$
begin{aligned}
frac{2 pi r}{v}&=2 pi sqrt{frac{r^{3}}{G m_{p}}}\
frac{1}{v}&=sqrt{frac{r}{G m_{p}}}/^{2}\
frac{1}{v^{2}}&=frac{r}{G m_{p}}\
m_{p}&=frac{rv^{2}}{G}
end{aligned}
$$
$$
begin{aligned}
frac{2 pi r}{v}&=2 pi sqrt{frac{r^{3}}{G m_{p}}}\
frac{1}{v}&=sqrt{frac{r}{G m_{p}}}/^{2}\
frac{1}{v^{2}}&=frac{r}{G m_{p}}\
m_{p}&=frac{rv^{2}}{G}
end{aligned}
$$
Step 4
4 of 5
We can substitute the values to find the mass of the planet
$$
begin{aligned}
m_{p}&=frac{rv^{2}}{G}\
m_{p}&=frac{6.7 cdot 10^{7} hspace{0.5mm} mathrm{m} cdot left( 2.0 cdot 10^{5} hspace{0.5mm} mathrm{m/s} right)^{2}}{6.7 cdot 10^{-11}hspace{0.5mm} mathrm{frac{m^{3}}{kg s^{2}}}}\
m_{p}&=4.0 cdot 10^{28} hspace{0.5mm} mathrm{kg}
end{aligned}
$$
$$
begin{aligned}
m_{p}&=frac{rv^{2}}{G}\
m_{p}&=frac{6.7 cdot 10^{7} hspace{0.5mm} mathrm{m} cdot left( 2.0 cdot 10^{5} hspace{0.5mm} mathrm{m/s} right)^{2}}{6.7 cdot 10^{-11}hspace{0.5mm} mathrm{frac{m^{3}}{kg s^{2}}}}\
m_{p}&=4.0 cdot 10^{28} hspace{0.5mm} mathrm{kg}
end{aligned}
$$
Result
5 of 5
$d)$ $4.0 cdot 10^{28} hspace{0.5mm} mathrm{kg}$
Exercise 3
Step 1
1 of 2
In this problem, we observing two satellites orbiting around the planet. To find their periods we use Newton’s version of Kepler’s third law. This law is the combination of Universal Gravitation and Kepler’s Third Law. We can write this law as
$$
begin{aligned}
T=2 pi sqrt{frac{r^{3}}{G m_{p}}}
end{aligned}
$$
where $T$ is a period of a satellite, $r$ is the radius of orbit of a satellite, $G$ is the gravitational constant, and $m_{S}$ is the mass of the planet.
$$
begin{aligned}
T=2 pi sqrt{frac{r^{3}}{G m_{p}}}
end{aligned}
$$
where $T$ is a period of a satellite, $r$ is the radius of orbit of a satellite, $G$ is the gravitational constant, and $m_{S}$ is the mass of the planet.
Step 2
2 of 2
As we can see, both of the satellites have the same orbital radius, and the mass of the planet is the same in both cases. Also, the period does not depend on the mass of the satellite, so it remains the same for both satellites.
We conclude that the correct statement is $A)$.
We conclude that the correct statement is $A)$.
Exercise 4
Step 1
1 of 3
In this problem we need to find the orbital period of the moon, so we use the equation
$$
begin{aligned}
T=frac{2 pi r}{v}
end{aligned}
$$
where $r$ is the orbital radius, and $v$ is the speed of the moon.
$$
begin{aligned}
T=frac{2 pi r}{v}
end{aligned}
$$
where $r$ is the orbital radius, and $v$ is the speed of the moon.
Step 2
2 of 3
We know the orbital radius, $r=5.4cdot 10^{7} hspace{0.5mm} mathrm{m}$, and the speed of the moon $v=9.0 cdot 10^{3} hspace{0.5mm} mathrm{m/s}$. Now, we can calculate the orbital period
$$
begin{aligned}
T&=frac{2 pi r}{v}\
T&=frac{2 pi 5.4cdot 10^{6} hspace{0.5mm} mathrm{m}}{9.0 cdot 10^{3} hspace{0.5mm} mathrm{m/s}}\
T&=1.16 picdot 10^{3} hspace{0.5mm} mathrm{s}
end{aligned}
$$
$$
begin{aligned}
T&=frac{2 pi r}{v}\
T&=frac{2 pi 5.4cdot 10^{6} hspace{0.5mm} mathrm{m}}{9.0 cdot 10^{3} hspace{0.5mm} mathrm{m/s}}\
T&=1.16 picdot 10^{3} hspace{0.5mm} mathrm{s}
end{aligned}
$$
Result
3 of 3
$c)$ $1.2pi cdot 10^{3} hspace{0.5mm} mathrm{s}$
Exercise 5
Step 1
1 of 7
In this problem, we need to find the relation between the gravitational forces from the Sun and from the planet. First, we are going to find the expressions for the forces, then we are going to divide them. Finally, we will calculate that relation. We use Universal Gravitational Law
$$
begin{aligned}
F=Gfrac{m_{1}m_{2}}{r^{2}}
end{aligned}
$$
where $m_{1}$ is the mass of the Sun or the planet, $m_{2}$ is the mass of the moon, and $r$ is the orbital radius.
$$
begin{aligned}
F=Gfrac{m_{1}m_{2}}{r^{2}}
end{aligned}
$$
where $m_{1}$ is the mass of the Sun or the planet, $m_{2}$ is the mass of the moon, and $r$ is the orbital radius.
Step 2
2 of 7
The gravitational force from the Sun is
$$
begin{aligned}
F_{S}=Gfrac{m_{S}m_{m}}{r_{S}^{2}}
end{aligned}
$$
$$
begin{aligned}
F_{S}=Gfrac{m_{S}m_{m}}{r_{S}^{2}}
end{aligned}
$$
Step 3
3 of 7
The gravitational force from the planet is
$$
begin{aligned}
F_{p}=Gfrac{m_{p}m_{m}}{r_{p}^{2}}
end{aligned}
$$
$$
begin{aligned}
F_{p}=Gfrac{m_{p}m_{m}}{r_{p}^{2}}
end{aligned}
$$
Step 4
4 of 7
Their relation is
$$
begin{aligned}
frac{F_{p}}{F_{S}}&=frac{Gfrac{m_{p}m_{m}}{r_{p}^{2}}}{Gfrac{m_{S}m_{m}}{r_{S}^{2}}}\
frac{F_{p}}{F_{S}}&=frac{frac{m_{p}}{r_{p}^{2}}}{frac{m_{S}}{r_{S}^{2}}}\
frac{F_{p}}{F_{S}}&=frac{m_{p}r_{S}^{2}}{m_{S}r_{p}^{2}}\
end{aligned}
$$
$$
begin{aligned}
frac{F_{p}}{F_{S}}&=frac{Gfrac{m_{p}m_{m}}{r_{p}^{2}}}{Gfrac{m_{S}m_{m}}{r_{S}^{2}}}\
frac{F_{p}}{F_{S}}&=frac{frac{m_{p}}{r_{p}^{2}}}{frac{m_{S}}{r_{S}^{2}}}\
frac{F_{p}}{F_{S}}&=frac{m_{p}r_{S}^{2}}{m_{S}r_{p}^{2}}\
end{aligned}
$$
Step 5
5 of 7
We know the mass of the Sun is $m_{S}=2.0 cdot 10^{30} hspace{0.5mm} mathrm{kg}$, and the mass of the planet is $m_{p}=2.4 cdot 10^{26} hspace{0.5mm} mathrm{kg}$. Also, the orbital radius from the moon to the Sun is $r_{S}=1.5 cdot 10^{11} hspace{0.5mm} mathrm{m}$, and the orbital radius from the moon to the planet is $r_{p}=6.0 cdot 10^{8} hspace{0.5mm} mathrm{m}$.
Step 6
6 of 7
Now, we can substitute the values
$$
begin{aligned}
frac{F_{p}}{F_{S}}&=frac{m_{p}r_{S}^{2}}{m_{S}r_{p}^{2}}\
frac{F_{p}}{F_{S}}&=frac{2.4 cdot 10^{26} hspace{0.5mm} mathrm{kg} cdot left(1.5 cdot 10^{11} hspace{0.5mm} mathrm{m} right)^{2}}{2.0 cdot 10^{30} hspace{0.5mm} mathrm{kg} cdot left( 6.0 cdot 10^{8} hspace{0.5mm} mathrm{m} right)^{2}}\
frac{F_{p}}{F_{S}}&=7.5
end{aligned}
$$
$$
begin{aligned}
frac{F_{p}}{F_{S}}&=frac{m_{p}r_{S}^{2}}{m_{S}r_{p}^{2}}\
frac{F_{p}}{F_{S}}&=frac{2.4 cdot 10^{26} hspace{0.5mm} mathrm{kg} cdot left(1.5 cdot 10^{11} hspace{0.5mm} mathrm{m} right)^{2}}{2.0 cdot 10^{30} hspace{0.5mm} mathrm{kg} cdot left( 6.0 cdot 10^{8} hspace{0.5mm} mathrm{m} right)^{2}}\
frac{F_{p}}{F_{S}}&=7.5
end{aligned}
$$
Result
7 of 7
$d)$ $7.5$
Exercise 6
Step 1
1 of 4
In this problem, we need to find the distance of the satellite from the planet. To do this, we are going to use Kepler’s third law, and we can write it as
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}=left( frac{r_{s2}}{r_{s1}}right)^{3}
end{aligned}
$$
where $T_{s1}$ is a period of the first satellite, $T_{s2}$ is a period of the second satellite, $r_{s1}$ is a radius of the first satellite’s orbit, and $r_{s2}$ is a radius of the second satellite’s orbit.
$$
begin{aligned}
left( frac{T_{s2}}{T_{s1}} right)^{2}=left( frac{r_{s2}}{r_{s1}}right)^{3}
end{aligned}
$$
where $T_{s1}$ is a period of the first satellite, $T_{s2}$ is a period of the second satellite, $r_{s1}$ is a radius of the first satellite’s orbit, and $r_{s2}$ is a radius of the second satellite’s orbit.
Step 2
2 of 4
We know that the period of the first satellite is $T_{s1}=20 hspace{0.5mm} mathrm{days}$, and its radius of orbit is $r_{s1}=2 cdot 10^{8} hspace{0.5mm} mathrm{m}$. The orbital period of the second satellite is $T_{s2}=160 hspace{0.5mm} mathrm{days}$.
Step 3
3 of 4
Now, we can find the period distance of the second satellite from the planet
$$
begin{aligned}
left( frac{r_{s2}}{r_{s1}}right)^{3}&=left( frac{T_{s2}}{T_{s1}} right)^{2}\
left( frac{r_{s2}}{r_{s1}}right)^{3}&=left( frac{160 hspace{0.5mm} mathrm{days}}{20 hspace{0.5mm} mathrm{days}} right)^{2}\
left( frac{r_{s2}}{r_{s1}}right)^{3}&=64\
frac{r_{s2}}{r_{s1}}&=4\
r_{s2}&=4r_{s1}\
r_{s2}&=4cdot 2 cdot 10^{8} hspace{0.5mm} mathrm{m}\
r_{s2}&=8 cdot 10^{8} hspace{0.5mm} mathrm{m}\
r_{s2}&=8 cdot 10^{5} hspace{0.5mm} mathrm{km}\
end{aligned}
$$
$$
begin{aligned}
left( frac{r_{s2}}{r_{s1}}right)^{3}&=left( frac{T_{s2}}{T_{s1}} right)^{2}\
left( frac{r_{s2}}{r_{s1}}right)^{3}&=left( frac{160 hspace{0.5mm} mathrm{days}}{20 hspace{0.5mm} mathrm{days}} right)^{2}\
left( frac{r_{s2}}{r_{s1}}right)^{3}&=64\
frac{r_{s2}}{r_{s1}}&=4\
r_{s2}&=4r_{s1}\
r_{s2}&=4cdot 2 cdot 10^{8} hspace{0.5mm} mathrm{m}\
r_{s2}&=8 cdot 10^{8} hspace{0.5mm} mathrm{m}\
r_{s2}&=8 cdot 10^{5} hspace{0.5mm} mathrm{km}\
end{aligned}
$$
Result
4 of 4
$r_{s2}=8 cdot 10^{5} hspace{0.5mm} mathrm{km}$
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