1302 Weight

$\begin{aligned} w &= F_g = \frac{Gm_E m}{R_E^2}\\ g &= \frac{Gm_E}{R_E^2} \end{aligned}$

Exercises

11, 12, 13, 15

(13.11) At what distance above the surface of the earth is the acceleration due to the earth's gravity $0.980 m/s^2$ if the accelera tiondue to gravity at the surface has magnitude $9.80 m/s^2$ ?

Solution

$(\sqrt {10} - 1) \cdot R_E = 1.38 \cdot 10^7m$

(13.12) The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth.
(a) Compute the acceleration due to gravity on the surface of Venus from these data.
(b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venus?

Solution

Todo

(13.13) Titania, the largest moon of the planet Uranus, has $\frac{1}{8}$ the radius of the earth and $\frac{1}{1700}$ the mass of the earth.
(a) What is the acceleration due to gravity at the surface of Titania?
(b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

Solution

a. The acceleration due to gravity on earth is $g = \frac{GM_E}{R_E^2} = 9.8m/s^2$ . On Titania

$\begin{aligned} g_T &= \frac{GM_T}{R_T^2} = \frac{\frac{1}{1700} GM_E}{(\frac{R_E}{8})^2}\\ &= \frac{GM_E}{R_E^2} \cdot \frac{64}{1700}\\ &= 9.8 \cdot \frac{64}{1700} = 0.37m/s^2 \end{aligned}$

b. The average density of Titania is

$\begin{aligned} \rho &= \frac{M_T}{V_T} = \frac{M_T}{\frac{4}{3}\pi R_T^3}\\ &= \frac{\frac{1}{1700} \cdot 5.97 \cdot 10^{24}}{\frac{4}{3}\pi \cdot (\frac{1}{8} \cdot 6.37 \cdot 10^6)^3} = 1660kg/m^3 \end{aligned}$

(13.15) Calculate the earth's gravity force on a 75-kg astronaut who is repairing the Hubble Space Telescope 600 km above the earth's surface, and then compare this value with his weight at the earth's surface. In view of your result, explain why it is said that astronauts are weightless when they orbit the earth in a satellite such as a space shuttle. Is it because the gravitational pull of the earth is negligibly small?

Solution

Todo