Problem#1
At what distance above the surface of the earth is the acceleration due to the earth’s gravity 0.980 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s2?Answer:
An object with mass m will experience a gravitational force equal to
Fg = GmME/r2
With r = distance of objects to the center of the earth
Then the gravitational acceleration of the object is given by
ag = Fg/m = GME/r2
Let a body of mass m be placed on the surface of the Earth (r = RE):
g = GME/RE2
Let the acceleration due to gravity at that position r = RE + h be g’ . Then,
g' = GME/(RE + h)2
For comparison, the ratio between g’ and g is taken
g'/g = [RE/(R + h)]2
0.980 m/s2/9.80 m/s2 = [6.4×106 m/(6.4 × 106 m + h)]2
0.316 (6.4 × 106 m + h) = 6.4 × 106 m
h = 1.38 x 107 m
Problem#2
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?
Answer:
Known:
Mass of Venus is 81.5% that of the earth, MV = 0.815ME
Radius of Venus is 94.9% that of the earth, RV = 0.949RE
(a) Let a body of mass m be placed on the surface of the Earth and Venus is
gE = GME/RE2 and gV = GMV/RV2
For comparison, the ratio between gV and gE is taken
gV/gE = [MV/ME][RE2/RV2]
gV/9.80 m/s2 = 0.815 x (1/0.949)2
gV = 8.87 m/s2
(b) If a rock weighs 75.0 N on earth, the mass of the rock is
m = WE/gE = 75.0 N/9.80 m/s2 = 7.65 kg
then, the weight of the rock on Venus is
Wv = mgV = 7.65 kg x 8.87 m/s2 = 67.88 N
Problem#3
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 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.)
Answer:
Known:
Mass of Titania is 1/1700 that of the Earth, mT = ME/1700
Radius of Titania is 1/8 that of the Earth, RT = RE/8
the average density of Earth is ρave, E = 5.5 g/cm3
(a) the acceleration due to gravity at the surface of Titania is
gT = GMT/RT2 = G(ME/1700)/(RE/8)2
gT = (64/1700)GME/RE2
gT = 0.369 m/s2
(b) the average density of Titania is
ρave,T = mT/VT = mT/(4πRT3/3)
ρave,T = (ME/1700)/[4π(RE/8)3/3]
ρave,T = 512/1700 x ρave, E = 1.66 g/cm3
Problem#4
Rhea, one of Saturn’s moons, has a radius of 765 km and an acceleration due to gravity of 0.278 m/s2 at its surface. Calculate its mass and average density.
Answer:
Known:
radius Rhea, RR = 765 km
gravity of Rhea, gR = 0.278 m/s2
the acceleration due to gravity at the surface of Rhea is
gR = GMR/RR2
0.278 m/s2 = (6.67 × 10−11 Nm2/kg2)mR/(765 x 103 m)2
mR = 2.44 x 1021 kg
and average density Rhea is
ρave,R = mR/VR = mR/[4π(RR)3/3]
ρave,R = 3 x 2.44 x 1021 kg/[4π x (765 x 103 m)3]
ρave,R = 1301 kg/m3
Problem#5
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 we say 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?
Answer:
the earth’s gravity force on a 75-kg astronaut who is repairing the Hubble Space Telescope 600 km above the earth’s surface is
F’g = GmME/(h + RE)2
F’g = (6.67 × 10−11 Nm2/kg2)(75 kg)(5.98 × 1024 kg)/(600 x 103 m + 6378 x 103 m)2
F’g = 614.36 N
compare this value with his weight at the earth’s surface, Fg is
Fg = mg = 735 N
The gravitational pull on the astronaut is therefore not negligible. However, both the
Hubble Space Telescope and the astronaut feel the same acceleration GmE/r2 and
therefore don’t accelerate relative to one another, but they also are also on the same orbit
above the earth and are thus constantly “free falling” and do not feel their weight.
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