All Solutions
Page 178: Section Review
$T = (2) (3.1416) sqrt{dfrac{(4.495e12)^3}{(6.67e-11)*(1.99e30)}}$
$$
T = 5.20 times 10^9 s
$$
5.20 times 10^9 s
$$
– Orbital radius: $r = 4.495 times 10^{12} mathrm{~m}$;
– Mass of the sun: $M = 1.99 times 10^{30} mathrm{~kg}$;
– Universal gravitational constant: $G = 6.67 times 10^{-11} ,frac{text{Nm}^2}{text{kg}^2}$;
**Required:**
– Time $T$ needed for the Neptune to orbit the sun;
$$T = sqrt{ left( frac{4 pi ^2}{G M} right) r^3 }$$
$$begin{align*}
T &= sqrt{ left( frac{4 pi ^2}{G M} right) r^3 } \
&= sqrt{ left( dfrac{4 pi ^2}{6.67 times 10^{-11} ,frac{text{Nm}^2}{text{kg}^2} cdot 1.99 times 10^{30} mathrm{~kg}} right) left( 4.495 times 10^{12} mathrm{~m} right)^3 } \
&= 5.2 times 10^{9} mathrm{~s} \
&approx 6 times 10^{4} mathrm{~days}
end{align*}$$
$$boxed{ T approx 6 times 10^{4} mathrm{~days} }$$
$$
F=dfrac{GMm}{R^{2}}
$$
If we compare that with the expression for weight that reads:
$$
begin{align*}
F_g&=mg\
dfrac{GMm}{R^{2}}&=mg
end{align*}
$$
We can very easily express $g$ as:
$$
g=dfrac{GM}{R^{2}}
$$
From the above expression we conclude that with constant mass, a decrease in radius causes a quadratic increase in $g$ which means if for example R decreases by two times g will increase by 4 times.
$g = dfrac{G m_E}{R^2}$
Where R is the radius of Earth. If R decreases, the gravitational acceleration will increase.
$F = dfrac{G m_1 m_2}{r^2} = dfrac{(6.67e-11)*(15)*(15)}{(0.35)^2} = 1.2 times 10^{-7} N$
The weight of one package is:
$W = m g = (15) (9.80) = 150 N$
The ratio is:
$$
ratio = dfrac{1.2 times 10^{-7}}{150} = 8.3 times 10^{-10}
$$
$$
begin{align*}
m_1&=m_2=15 mathrm{kg}\
R&=35 mathrm{cm}=0.35 mathrm{m}\
G&=6.67cdot10^{-11} mathrm{dfrac{Nm^{2}}{kg^{2}}}
end{align*}
$$
Unknown: $F=?$, $dfrac{F}{F_g}=$?
$$
F=dfrac{Gm_1m_2}{R^2}
$$
When we insert known values into an expression and calculate it, we get:
$$
begin{align*}
F&=dfrac{6.67cdot10^{-11} mathrm{dfrac{Nm^{2}}{kg^{2}}15 mathrm{kg}15 mathrm{kg}}}{0.35^2 mathrm{m^2}}\
&=1.225times10^{-7} mathrm{N}
end{align*}
$$
We then calculate the weight of one package using:
$$
begin{align*}
F_g&=mg\
&=15 mathrm{kg}cdot 9.81 mathrm{dfrac{m}{s^2}}\
&=1.472 mathrm{N}
end{align*}
$$
to calculate what fraction of weight of one package is gravitational force, we simply divide these two forces, which give us ratio:
$$
begin{align*}
dfrac{F}{F_g}&=dfrac{1.472 mathrm{N}}{1.225times10^{-7} mathrm{N}}\
&=8.322times10^{-8}
end{align*}
$$
Which is a negligible value. Note that the units of measure cancel out and the end result is just a number.
begin{align*}
F&=1.225times10^{-7} mathrm{N}\
dfrac{F}{F_g}&=8.322times10^{-8}
end{align*}
$$
$$
begin{align*}
G&=frac{Fcdot{r}}{m_1cdot{m_2}}
end{align*}
$$
Where the $F$ represents the force of attraction between two bodies, $r$ is the distance between the centers of the bodies, and $m_1$ and $m_2$ stands for the masses of the bodies.
So, if we replace the body with the body of the same mass, the universal gas constant does not change.
$$
boxed{text{It is just the law, not the theory.}}
$$
In a horizontally launched projectile motion, the time needed for the projectile to hit the ground depends on the height difference. There are two components of motion:
First, the vertical component of motion is accelerated by the gravitational force and can be written as:
$$
begin{align}
y&=frac{1}{2}cdot{g}cdot{t^2}
end{align}
$$
The second is the horizontal component that is uniform with initial velocity $v_0$:
$$
begin{align}
x&=v_0cdot{t}
end{align}
$$
As we already said, the complete trajectory depends on the vertical distance that the projectile can pass, the horizontal component or the maximal distance of the projectile is the function of time needed for the drop. Let’s express the time of moving from the first equation.
$$
begin{align*}
t&=sqrtfrac{2cdot{y}}{g}
end{align*}
$$
If the gravitational force decrease and the mass of the rock are constant, the time needed for the rock to hit the ground will be increased. Also, the maximal reach of the projectile will be increased.
$$
boxed{text{The maximal reaching distance will be increased.}}
$$
$bold{b)}$
The force that acts on a foot in case of the accidental drop will be:
$$
begin{align*}
F=gcdot{m}
end{align*}
$$
In the case of lower gravity, the force will be lower so the pain will be less.
$$
boxed{text{Force will be lower and the man will fill less pain.}}
$$
b) Force will be lower and the man will fill less pain.
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