Calculating the force of attraction between the human and Earth’s center of gravity requires G, (a) Earth’s radius, and the masses of the two bodies. As the human is standing on the surface of the Earth, the distance between their centers of mass is simply Earth’s radius (ignoring the height of the human)
a. Calculating $F_g$ between the two masses at the surface, the distance between the two masses is the radius of the planet.
| 3 sf | $F_g = G \frac{m \cdot M}{r^2}$ |
|---|---|
| $m = 70.0 kg$ $M = 5.970x10^{24} kg$ $r = 6\,378x10^3 m$ |
$F_g =G \left(\frac{(70.0 kg)(5.970x10^{24}kg)}{(6\,378x10^{3} m)^2} \right)$ $F_g = 685.66 N$ $F_g \approx 685 N$ |
b. Since the gravitational force and the mass of the human are known, the acceleration can be calculated using Newton’s Second Law: The acceleration of the human is centripetal, toward Earth’s center.
| 3 sf | $F_g = m \cdot a_g$ |
|---|---|
| $m = 70.0 kg$ $F_g = 685.66 N$ $r = 6\,378x10^3 m$ |
$a_g = \frac{F_g}{m}$ $a_g = \frac{685.66 N}{70.0 kg}$ $a_g = 9.795142857… \frac{m}{s^2}$ $a_g = 9.80 \frac{m}{s^2}$ |
As the human is standing on the surface of the Europa, the distance between their centers of mass is simply Europa’s radius (ignoring the trivially small height of the human):
a. Calculating $F_g$ between the two masses at the surface, the distance between the two masses is the radius of the planet.
| 3 sf | $F_g = G \frac{m \dot M}{r^2}$ |
|---|---|
| $m = 70.0 kg$ $M = 4.897x10^{22} kg$ $r = 1\,561x10^3 m$ |
$F_g =G \left(\frac{(70.0 kg)(4.897x10^{22}kg)}{(1\,561x10^{3} m)^2} \right)$ $F_g = 113.38 N$ $F_g = 113 N$ |
b. Since the gravitational force and the mass of the human are known, the acceleration can be calculated using Newton’s Second Law: The acceleration of the human is centripetal, toward Europa’s center.
| 3 sf | $F_g = m \cdot a_g$ |
|---|---|
| $m = 70.0 kg$ $F_g = 113.38 N$ $r = 1\,561x10^3 m$ |
$a_g = \frac{F_g}{m}$ $a_g = \frac{113.38 N}{70.0 kg}$ $a_g = 1.6197142857… \frac{m}{s^2}$ $a_g = 1.62 \frac{m}{s^2}$ |
c. The acceleration of Luna can be calculated using the known force of gravity and the mass of Luna:
| 2 sf | $F_g = m \cdot a_g$ |
|---|---|
| $m_{Luna} = 0.073 \times 10^{24} kg$ $F_{g_{Luna}} = 113.38 N$ $r = 1\,561x10^3 m$ |
$a_{g_{Luna}} = \frac{F_g}{m}$ $a_{g_{Luna}} = \frac{113.38 N}{0.642 \times 10^{24} kg}$ $a_{g_{Luna}} = 1.547… \times 10^{-21} \frac{m}{s^2}$ $a_g = 1.5 \times 10^{-21} \frac{m}{s^2}$ |
a. On the surface of Earth
| 3 sf | $F_g = m \cdot a_g$ |
|---|---|
| $m=45.5kg$ $a_g=9.80 \frac{m}{s^2}$ |
$F_g=45.5 kg \cdot 9.80 \frac{m}{s^2}$ $F_g=446N$ |
b. At a distance of $1.0 x 10^3 km$ above Earth:
At that altitude, gravity would act over a distance of $r = r_E + 1.0 \times 10^6 \, m = 7.378 \times 10^6 \, m$
| 3 sf | $F_g = G \frac{m \cdot M_{E}}{r^2}$ |
|---|---|
| $m = 45.5$ \, kg $M_E = 5.972 \times 10^{24} , kg$ $r = 7.371 \times 10^6 m$ |
$F_g = G \frac{(45.5 kg)(5.972 \times 10^{24} kg)}{(7.378 \times 10^6)^2}$ $F_g = 333 \,N$ |
c. On the “surface” of Saturn
| 3 sf | $F_g = G \frac{m\cdot M_{Saturn}}{r^2}$ |
|---|---|
| $m = 45.5 \, kg$ $M_{Saturn} = 5.683 \times 10^{26} \, kg$ $r_{Saturn} = 60 \,268 \times 10^3 m$ |
$F_g = G \frac{(45.5 kg)(5.683 \times 10^{26} \,kg)}{(60 \,268 \times 10^3 m)^2}$ $F_g = 475 \, N$ |
d. Values from (a) and (c): The values are similar because the ratio of mass to the square of the radius, $\frac{M}{r^2}$, which determines the acceleration due to gravity is somewhat comparable for Earth’s surface and Saturn’s cloud tops. Saturn is approximately 100x the mass of Earth, but it is also approximately 10x larger in radius:
Since $ F_g \propto \frac{M}{r^2}$, and $\frac {M_{Saturn}}{r_{Saturn}^2} \approx \frac{100 \cdot M_{Earth}}{(10\cdot r_{Earth})^2}$ = $\frac{\cancel{100} \cdot M}{\cancel{100} \cdot r^2}$
so the forces should be approximately equal.
e. $1.0 \times 10^3 km$ above Saturn’s atmosphere.
At that altitude, gravity would act over a distance of $r = r_S + 1.0 \times 10^6 \, m = 6.1268 \times 10^7 \,m$
| 3 sf | $F_g = G \frac{m \cdot M_Saturn}{r^2}$ |
|---|---|
| $m = 45.5 , kg$ $M_{Saturn} = 5.68 \times 10^{26} \, kg$ $r = 6.1268 \times 10^7 \,m$ |
$F_g = G \frac{(45.5\,kg)(5.683 \times 10^{26} \, kg)}{(6.1268 \times 10^7)^2}$ $F_g = 459.66… \, N$ $F_g = 4.60 \times 10^2\, N$ |
f. An altitude of $1.0x10^6 m$ on Earth represents a change of $\approx$ 15% of Earth’s radius. In contrast, $1.0 \times 10^6 \, m$ above Saturn’s outermost clouds represents only $\approx$ 1.5% change in the radius. Thus, rising 1 000 km above Earth will affect the force of gravity much more substantially than rising 1 000 km above Saturn’s much larger radius.
A spacecraft orbiting 2 Earth radii above the surface would experience gravity acting over a total distance of $3 \times r_E$ or $(3)(6\,378 \times 10^3 m)$ which is $6.378 \times 10^6 m$
| 3 sf | $F_g = G \frac{m_s M_E}{r^2}$ |
|---|---|
| $m_s = 1\,850 \, kg$ $M_E = 5.972 \times 10^{24} , kg$ $r =6.378 \times 10^6 m$ |
$F_g = G \frac{(1\,850 \,kg)(5.972 \times 10^{24} \, kg)}{(6.378 \times 10^6 m)^2}$ $F_g = 2\,010 \, N$ |
| 3 sf | $F_{net} = ma$ |
|---|---|
| $F_{net} = F_g = 2\,010 , N$ $m_s = 1\,850 , kg$ |
$a_c = \frac{F_{net}}{m_s}$ $a_c = \frac{2010 \, N}{1850 , kg}$ $a_c = 1.09 \, m/s^2$ |
| 3 sf | $a_c = \frac{v^2}{r}$ |
|---|---|
| $a_c = 1.09 , m/s^2$ $r = 1.9113 \times 10^7 , m$ |
$v^2 = a_c r$ $v = \sqrt{a_c r}$ $v = \sqrt{(1.09)(1.9113 \times 10^7)}$ $v = \sqrt{2.083317 \times 10^7}$ $v = 4\,560 \, m/s$ |
Problem 6: Hypothetical Planet
This problem can be solved computationally using Newton’s Universal Law of Gravitation (see below) or conceptually using the inverse square law:
Let $P$ represent the hypothetical planet:
$F \propto \frac{1}{r^2}$
$m_E \cdot a_{g_{Earth}} = \frac{1}{(r_{Earth})^2} = 9.80 \frac{m}{s^2}$
$m_P \cdot a_{g_{P}} = \frac{1}{(2 \cdot r_{Earth})^2} = \frac{1}{4 \cdot (r_{Earth})^2}$
and since $m_E = m_P$:
$a_{g_{Earth}} \propto \frac{1}{4} \cdot a_{g_P} = (\frac{1}{4})(9.80 \frac{m}{s^2}) \approx 2.45 \frac{m}{s^2}$
Alternately, use Universal Gravitation:
| 3 sf | $a_{g_P} = G \frac{M_P}{r^2}$ |
|---|---|
| $M_P = M_E$ $R_P = 2.0 R_E$ |
$a_{g_{P}} = G \frac{M_E}{(2.0 R_E)^2} = \frac{1}{4} G \frac{M_E}{R_E^2}$ $a_{g_{P}} = \frac{1}{4} a_g = \frac{1}{4} (9.80)$ $a_{g_{P}} = 2.45 \, m/s^2$ |
Problem 7: Lander Probe on Europa
a. Determine the force of gravity on Europa’s surface.
| 3 sf | (F_g = m_{Lander} \cdot a_{g_{Europa}}) |
|---|---|
| $m_{Lander} = 958 \, kg$ $a_{g_{Europa}} = 1.98 \, \frac{m}{s^2}$ |
$F_g = (958 \, kg)(1.98\, \frac{m}{s^2})$ $F_g = 1897 \, N$ $F_g =1.90 x 10^3 N$ |
b. Calculating the circumference of Europa first requires knowing its radius, $r$. Note that this could also be determined using $F_g = G \frac{M \cdot m}{r^2}$
| 3 sf | $a_{g_{Europa}} = G \frac{M_{Europa}}{r_{Europa}^2}$ |
|---|---|
| $a_{g_{Europa}} = 1.98 \, m/s^2$ $M_{Europa} = 4.9 \times 10^{22} \, kg$ |
$r_{Europa} = \sqrt{\frac{G \cdot M_{Europa}}{a_{g_{Europa}}}} $ $r_{Europa} = \sqrt{\frac{G(4.9 \times 10^{22} \,kg)}{1.98 \, \frac{m}{s^2}}}$ $r_{Europa} = 1.29 \times 10^6 \, m$ |
| 3 sf | $C = 2 \pi r$ |
|---|---|
| $r_{Europa} = 1.29 \times 10^6 \, m$ | $C = 2 \pi (1.29 \times 10^6)$ $C = 8.09 \times 10^6 \, m$ $C = 8\,090 \, km$ |
Note: using the unrounded value for Europa’s radius gives a circumference closer to $8.07x10^6 \, m$
Problem 8: Jupiter and Saturn Attraction
| 3 sf | $F_g = G \frac{m_1 m_2}{r^2}$ |
|---|---|
| $M_J = 1.898 \times 10^{27} \, kg$ $M_S = 5.683 \times 10^{26} \, kg$ $r = 6.4627 \times 10^{11} \, m$ $G = 6.673 \times 10^{-11}$ |
$F_g = G \frac{(1.898 \times 10^{27})(5.683 \times 10^{26})}{(6.4627 \times 10^{11})^2}$ $F_g = G \frac{1.0786 \times 10^{54}}{4.1766 \times 10^{23}}$ $F_g = 1.72 \times 10^{20} \, N$ |