Problem Set 5.9:  25
 27 
29  31  39  41

43  45  49  51
 52  57  63  67
 68  75  78  Go up
 by Cooper Spear and Chris Murray, 2001
25. Calculate the force of gravity on a spacecraft 12,800 km ( 2 earth radii) above the Earth's surface if its mass is 1400 kg?
Earth has a radius of 6.38x10^{6} m, and a mass of 5.98x10^{24} kg, now since the spacecraft is two earth radii above the surface of the earth, it is three earth radii from the center, which is the distance you want to use in the formula for gravity. So now we know that the center to center distance is 3(6.38x10^{6 }m) = 1.914x10^{7} m
So now we just plug into the formula for gravitational force:
F = G(m_{1}m_{2})/(r^{2})
F = (6.67x10^{11} Nm^{2}/kg^{2})(5.98x10^{24} kg)(1400 kg)/(1.914x10^{7} m)^{2} = 1.5243x10^{3 }N = 1.52x10^{3 }N
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27. A hypothetical planet has a radius 2.5 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
We don't really have an IB formula for the acceleration of gravity on the surface of a planet, but we know that F = ma, so the acceleration of an object with mass m is: a = F/m. Where F is the force of gravity given by F = G(m_{1}m_{2})/(r^{2}). Where m_{1} might be the planet's mass, and m_{2} is the mass of the object.
Now we just plug the formula for the force of gravity into Newton's second law and we get the acceleration of gravity at the planet's surface:
a = F/m = {F = G(m_{1}m_{2})/(r^{2})}/m_{2 }= G(m_{1})/(r^{2}) so the acceleration of gravity at the surface is:
a = G(m_{1})/(r^{2})
I will use 6.38x10^{6}m as the radius of the earth, so this planet has a radius of 2.5(6.38x10^{6}m) = 1.5950x10^{7}m
Earth has a mass of 5.98x10^{24}kg, and so let's just plug this into our equation:
a = (6.67x10^{11} Nm^{2}/kg^{2})(5.98x10^{24} kg)/(1.5950x10^{7}m)^{2} = 1.5679 m/s/s = 1.6 m/s/s
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29. At the surface of a certain planet, the gravitational acceleration g has a magnitude of 12.0 m/s². A 2.10kg brass ball is transported to this planet. What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?
Well, uh gee Dougger, the mass is universal. 2.10 kg of brass is 2.10 kg of brass no matter where it is. The weight, however, depends on the strength of the gravitational field the ball is in. On earth the field strength near the surface is 9.8 m/s/s or 9.8 N/kg. On this planet it is 12.0 m/s/s or 12.0 N/kg.
So on Earth, the ball weighs:
F = ma = (2.10 kg)(9.80 N/kg) = 20.58 N = 20.6 N
So on Planet Zirkon, the ball weighs:
F = ma = (2.10 kg)(12.0 N/kg) = 25.2 N
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31. An exotic finish to massive stars is that of a neutron star, which might have as much as five times the mass of our Sun packed into s sphere about 10 km in radius!!!!!! Estimate the surface gravity on this monster.
We don't really have an IB formula for the acceleration of gravity on the surface of a sphere, but we know that F = ma, so the acceleration of an object with mass m is: a = F/m. Where F is the force of gravity given by F = G(m_{1}m_{2})/(r^{2}). Where m_{1} might be the sphere's mass, and m_{2} is the mass of the object.
Now we just plug the formula for the force of gravity into Newton's second law and we get the acceleration of gravity at the neutron star's surface:
a = F/m = {F = G(m_{1}m_{2})/(r^{2})}/m_{2 }= G(m_{1})/(r^{2}) so the acceleration of gravity at the surface is:
a = G(m_{1})/(r^{2})
The sun has a mass of 1.99x10^{30}kg so this neutron star has a mass five times that: 5(1.99x10^{30}kg) = 9.95x10^{30}kg
Its radius is 10 km or 10,000 m
Now we just plug into our formula
a = G(m_{1})/(r^{2})
a = (6.67x10^{11} Nm^{2}/kg^{2})(9.95x10^{30}kg)/(10,000m)^{2} = 6.6367x10^{12} m/s/s = 7x10^{12} m/s/s
A tough place to live. Holy gravitational red shift, Batman.
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39. Calculate the velocity of a satellite moving in a stable circular orbit about the Earth at a height of 3600 km?
3600 km = 3,600,000 m = 3.6x10^{6} m
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m_{1}m_{2})/(r^{2}), G = 6.67 x 10^{11} Nm^{2}/kg^{2}
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m_{1}m_{2})/(r^{2}) = ma  where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
or
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
(For both of these, the satellite is m_{1} because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
Now, for the satellite orbiting the earth we know the other mass (m_{2}) is the earth's mass (5.98 x 10^{24} kg) and the satellite mass is unknown (m_{1}). Since we want to know the velocity, let's use the orbital condition that has velocity in it:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
Now, I could just put in numbers, but as these are in scientific notation, and sometimes rather large, I would rather do math with symbols, so let's solve for v:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r  (divided both sides by m_{1})  the satellite mass cancels  that is why we don't need to know it)
G(m_{2})/(r^{2}) = v^{2}/r  (multiply both sides by r)
G(m_{2})/(r) = v^{2}
Finally,
v = Ö(Gm_{2}/r).
There is one last tricky thing about this problem, and that is we are not given the orbital radius which we need to know to solve the problem. We are given the elevation above the earth's surface. To get the distance to the center of the earth, which is what we want for a circular orbit, we need to add the radius of the earth (6.38x10^{6}m). so
r = 6.38x10^{6 }m + 3.6x10^{6} m = 9.98 x 10^{6} m, so we are ready to plug and chug:
v = Ö(Gm_{2}/r) = Ö{( 6.67 x 10^{11} Nm^{2}/kg^{2})(5.98 x 10^{24} kg)/(9.98 x 10^{6})} = 6321.9 m/s= 6300 m/s
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41. At what rate must the cylindrical spaceship of Fig. 530 rotate (see Question 12), if occupants are to experience simulated gravity of 0.5 g? Assume the spaceship's diameter is 32 m, and give your answer as the time needed for one revolution.
If the diameter is 32 m, then the radius is 16 m
Well, .5 g is .5(9.8 m/s/s) = 4.9 m/s/s. This is what you want the centripetal acceleration to be it you want a half a "g" of artificial gravity. All we need to do is to find the period of revolution with the given radius, and 4.9 m/s/s as the acceleration:
a = (4π^{2}r)/T^{2 }
Now, just solve for T:
aT^{2 } = (4π^{2}r)
T^{2 } = (4π^{2}r)/a
And finally,
T = Ö{(4π^{2}r)/a}
So let's plug the numbers in:
T = Ö{4π^{2}(16 m)/(4.9 m/s/s)} = 11.35 s = 11 s
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43. During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 100 km. How long did it take to go around the Moon once?
100 km = 100,000 m
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m_{1}m_{2})/(r^{2}), G = 6.67 x 10^{11} Nm^{2}/kg^{2}
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m_{1}m_{2})/(r^{2}) = ma  where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
or
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
(For both of these, the satellite is m_{1} because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
Now, for you orbiting the moon we know the other mass (m_{2}) is the moon's mass (7.36 x 10^{22} kg) and your mass is unknown (m_{1}). Since we want to know the period of motion, let's solve the condition of orbit with the period in it:
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
Solving for the period, T:
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }Cross multiply^{ } G(m_{1}m_{2})T^{2} = m_{1}(4π^{2}r)(r^{2})
Divide by m_{1} and combine r:
G(m_{2})T^{2} = 4π^{2}r^{3}
Divide both sides by Gm_{2}:
T^{2} = 4π^{2}r^{3}/Gm_{2}
Finally:
T = Ö{4π^{2}r^{3}/Gm_{2}} = Ö{4π^{2}(1.838x10^{6}m)^{3}/(6.67 x 10^{11} Nm^{2}/kg^{2})(7.36 x 10^{22} kg)} = 7066.37 s
T = (7066.37 s)/(3600 s/hr) = 1.96 hr = 2.0 hr
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45. The rings of Saturn are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73,000 km, while the outer radius is 170,000 km. Find the period of an orbiting chunk at the inner radius and the period of a chunk at the outer radius, compare your numbers with Saturn's mean rotation period of 10 hours and 39 minutes. The mass of Saturn is 5.69 x 10^{26} kg.
73,000 km = 7.3x10^{7} m
170,000 km = 1.7x10^{8} m
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m_{1}m_{2})/(r^{2}), G = 6.67 x 10^{11} Nm^{2}/kg^{2}
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m_{1}m_{2})/(r^{2}) = ma  where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
or
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
(For both of these, the satellite is m_{1} because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
Now, for something orbiting Saturn we know the other mass (m_{2}) is Saturn's mass (5.69 x 10^{26} kg) and the satellite mass is unknown (m_{1}). Since we want to know the period of motion, let's solve the condition of orbit with the period in it:
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
Solving for the period, T:
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }Cross multiply^{ } G(m_{1}m_{2})T^{2} = m_{1}(4π^{2}r)(r^{2})
Divide by m_{1} and combine r:
G(m_{2})T^{2} = 4π^{2}r^{3}
Divide both sides by Gm_{2}:
T^{2} = 4π^{2}r^{3}/Gm_{2}
Finally:
T = Ö{4π^{2}r^{3}/Gm_{2}}
So let's plug in for the inner ring with radius 7.3x10^{7} m and Saturn's mass of 5.69x10^{26 }kg:
T = Ö{4π^{2}(7.3x10^{7} m)^{3}/( 6.67 x 10^{11} Nm^{2}/kg^{2})(5.69x10^{26 }kg)} = 20116.16 s
T = (20116.16 s)/(3600 s/hr) = 5.5878 hr
Since there are 60 minutes in an hour, .5878 hr = (.5878 hr)(60min/hr) = 35 minutes, so instead of 5.5878 hrs, it is really
T = 5 hours, 35 minutes
Now let's plug in for the outer ring with radius 1.7x10^{8} m and Saturn's mass of 5.69x10^{26 }kg:
T = Ö{4π^{2}(1.7x10^{8} m)^{3}/(6.67 x 10^{11} Nm^{2}/kg^{2})(5.69x10^{26 }kg)} = 71488.17 s
T = (71488.17 s)/(3600 s/hr) = 19.8578 hr
Since there are 60 minutes in an hour, .8578 hr = (.8578 hr)(60min/hr) = 51 minutes, so instead of 19.8578 hrs, it is really
T = 19 hours, 51 minutes
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49. Suppose that a binary star system consists of two stars of equal mass. They are observed to be separated by 360 million km and to take 5.0 Earth years to orbit about a point midway between them. What is the mass of each?
360 million km = 360x10^{6} km = 3.60x10^{11} m
5.0 Earth years = (5.0 yr)(3.156x10^{7} s/yr) = 1.578x10^{8} s
This is not a normal orbit problem, but it is not much different. The key to this problem is to realize that the planets are orbiting a point midway between them. Their radius of orbit is half their separation. Since they are 3.60x10^{11} m apart, they move in a circle 1.80x10^{11} m in radius. So now, we have gravity providing the centripetal force for them to orbit each other, so we set the force of gravity equal to the centripetal force needed for them to orbit the center. The only hitch is that if "r" is the radius of orbit, the center to center distance is now 2r. So let's set up the orbital condition with the period in it:
G(m_{1}m_{2})/(2r)^{2} = m_{1}(4π^{2}r)/T^{2 }Since both masses are the same, I will drop the subscripts:^{ }Gm^{2}/(2r)^{2} = m4π^{2}r/T^{2}_{ }^{ } Canceling m:
Gm/(2r)^{2} = 4π^{2}r/T^{2}_{ }^{ } Cross multiplying:
GmT^{2} = 4π^{2}r(2r)^{2} ^{ }=^{ }16π^{2}r^{3} Dividing both sides by GT^{2}:
m = 16π^{2}r^{3}/{GT^{2}} = 16π^{2}(1.80x10^{11} m)^{3}/{(6.67 x 10^{11} Nm^{2}/kg^{2})(1.578x10^{8} s)^{2}} = 5.54495x10^{29} kg = 5.5x10^{29} kg
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51. Use Kepler's laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the Earth's surface.
Well, the moon is 3.84x10^{8} m from the center of the earth, and the satellite is 6.38x10^{6} m, so we will use Kepler's law of periods and radii:
T_{1}^{2}/R_{1}^{3} = T_{2}^{2}/R_{2}^{3 }Let's make
T_{1} = the moon's period = 27.4 d
R_{1} = the moon's radius of orbit = 3.84x10^{8} m
R_{2} = The radius of the satellite's orbit = 6.38x10^{6} m^{ }T_{2} = The period of the satellite^{ }Now let's solve Kepler's equation for T_{2}:^{ }T_{1}^{2}/R_{1}^{3} = T_{2}^{2}/R_{2}^{3 }T_{1}^{2}R_{2}^{3}/R_{1}^{3} = T_{2}^{2}
Ö{(27.4 d)^{2}(6.38x10^{6} m)^{3}/(3.84x10^{8} m)^{3}} = T_{2}^{ }Ö{(27.4 d)^{2}(6.38x10^{6} m)^{3}/(3.84x10^{8} m)^{3}} = 0.058679276 day = (0.058679276 day)(24 hr/day) = 1.41 hrs
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52. The asteroid Icarus, though only a few hundred meters across, orbits the Sun like the other planets. Its period is about 410 d. What is its mean distance from the sun?
Well, the Earth has an orbital radius of 1.50x10^{11} m and a period of 365 days and Icarus has a period of 410 days, so we will use Kepler's law of periods and radii:
T_{1}^{2}/R_{1}^{3} = T_{2}^{2}/R_{2}^{3 }Let's make
T_{1} = the Earth's period = 365 d
R_{1} = the Earth's radius of orbit = 1.50x10^{11} m
T_{2} = The period of Icarus = 410 d
R_{2} = The radius of Icarus's orbit (the unknown)^{ }Now let's solve Kepler's equation for R_{2}:^{ }T_{1}^{2}/R_{1}^{3} = T_{2}^{2}/R_{2}^{3 }T_{1}^{2}R_{2}^{3} = T_{2}^{2}R_{1}^{3}
R_{2}^{3} = T_{2}^{2}R_{1}^{3}/T_{1}^{2}
R_{2} = ^{3}Ö{T_{2}^{2}R_{1}^{3}/T_{1}^{2}}
R_{2} = ^{3}Ö{(410 d)^{2}(1.50x10^{11} m)^{3}/(365 d)^{2}} = 1.62x10^{11} m^{ }(Table of contents)
57. Table 52 gives the mean distance, period, and mass for the four largest moons of Jupiter (those discovered by Galileo in 1609). (a) Determine the mass of Jupiter using the data for Io. (b) Determine the mass of Jupiter using data for each of the other three moons. Are the results consistent?
Let's use the orbital condition (see problem 45) that has the period in it, and solve for the mass of the planet:
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }G(m_{2})/(r^{2}) = (4π^{2}r)/T^{2 }G(m_{2}) = (4π^{2}r)(r^{2})/T^{2 }G(m_{2}) = 4π^{2}r^{3}/T^{2 }m_{2} = 4π^{2}r^{3}/GT^{2 }So for Io, r = 422x10^{3} km = 4.22x10^{8} m, and T = 1.77 days = (1.77 d)(24 h/d)(3600 s/hr) = 152928 s. Plugging into our formula:
m_{2} = 4π^{2}r^{3}/GT^{2} = 4π^{2}(4.22x10^{8} m)^{3}/(6.67 x 10^{11} Nm^{2}/kg^{2})(152928 s)^{2 }= 1.90x10^{27} kg
Applying the same formula to the data from the other planets, I get the following results: (Be sure to convert radii to meters, and period to seconds)
Moon
Period (d)
Period (s)Orbital Radius (m) Mass of Jupiter (kg) Io 1.77 152928 4.22x10^{8} 1.90x10^{27} Europa 3.55 306720 6.71x10^{8} 1.90x10^{27} Ganymede 7.16 618624 1.07x10^{8} 1.89x10^{27} Callisto 16.7 1442880 1.88x10^{9} 1.90x10^{27} These masses of Jupiter are pretty close to each other, and that's because it is the same Jupiter. (Meant to be funny)
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63. Tarzan plans to cross a gorge by swinging in an arc from a hanging vine (Fig. 541). If his arms are capable of exerting a force of 1400 N on the rope, what is the maximum speed he can tolerate at the lowest point of his weight? His mass is 80 kg and the vine is 4.8 m long.
Tarzan has a mass of 80 kg which means he weighs (80 kg)(9.8 N/kg) = 784 N. At the bottom of the arc, with 1400 N of force up (+) and 784 N of force down(), his F = ma looks like:
<1400N  784 N> = 80a
a = 7.7 m/s/s. This can be any upward acceleration, either linear or centripetal, and when he is swinging, it is centripetal, so using the equation for centripetal acceleration, we can solve for the velocity:
a = v^{2}/r
ar = v^{2 }v = Ö(ar) = Ö{(7.7 m/s/s)(4.8 m)} = 6.1 m/s
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65. On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every three seconds. If we assume their arms are each 0.80 m long, how hard are they pulling on one another, assuming their individual masses are 60.0 kg?
Each skater is giong in a circle with a radius of .80 m, with a period of 3.0 s, and has a mass of 60.0 kg, so the centripetal force needed would be:
F = m4π^{2}r/T^{2 } = (60 kg)4π^{2}(.80 m)/(3.0 s)^{2} = 210.6 N = 210 N
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67. At what distance from the Earth will a spacecraft traveling directly from the Earth to the Moon experience zero net force because the Earth and Moon pull with equal and opposite forces?
Let x be the distance from the Earth for this spot (It would be closer to the moon than the Earth), and let r be the Earth moon distance of 3.84x10^{8} m. The distance to the Moon is then (r  x) We also know the mass of the moon is 7.36x10^{22} kg, and the mass of the Earth is 5.98x10^{24} kg
At this point, the force of gravity is equal between the moon and the spacecraft, and between the Earth and the spacecraft, so we will set the force of gravity equal:
G(m_{E}m_{s})/(r_{E}^{2}) = G(m_{M}m_{s})/(r_{M}^{2})
Now we plug in r  x for the distance from the moon, and x for the distance from the Earth:
G(m_{E}m_{s})/(x^{2}) = G(m_{M}m_{s})/((rx)^{2}) and now solve for x:
(m_{E}m_{s})/(x^{2}) = (m_{M}m_{s})/((rx)^{2})
m_{E}/x^{2} = m_{M}/(rx)^{2 }Öm_{E}/x = Öm_{M}/(rx)  and finally cross multiply
Öm_{E}(rx) = Öm_{M}x
Öm_{E}rÖm_{E}x = Öm_{M}x
Öm_{E}r = Öm_{M}x + Öm_{E}x = x(Öm_{M} + Öm_{E})
x = Öm_{E}r/(Öm_{E} + Öm_{M}) = Ö(5.98x10^{24} kg)(3.84x10^{8} m)/{Ö(5.98x10^{24} kg) + Ö(7.36x10^{22} kg)}= 3.4565x10^{8} m = 3.46x10^{8} m
(From the center of the Earth, or 3.39x10^{8} m from the surface)
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68. A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire) (Fig. 542). The circle formed by the tube has a diameter of about 1.1 km. (a) On which wall inside the tube will people be able to walk? (b) What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the Earth (1.0 g) is to be felt?
If the diameter is 1.1 km or 1100 m, then the radius is 550 m
The people would have their heads toward the center, and would walk on the inside of the tube. (They are accelerating toward the center, so they feel their "g" force the opposite way, which is toward the outside. So "down" for them is toward the outside)
If you want 1 "g" of acceleration, then you would set the centripetal acceleration equal to 9.8 m/s/s:
g = (4π^{2}r)/T^{2 }
T^{2} = (4π^{2}r)/g
T = Ö(4π^{2}r/g) = 2πÖ(r/g) = 2πÖ(550 m/9.8 m/s/s) = 47.0704 sec/rev
Now this is sec per revolutions, and they asked for rev/day, so there are (24hr/day)(3600sec/hr) = 86400 sec/day, so
rev/day = (86400 sec/day,)/(47.0704 sec/rev) = 1835.547245 = 1800 rev/day
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75. Two equalmass stars maintain a constant distance apart of 8.0 x 10^{10} m and rotate about a point midway between them at a rate of one revolution every 12.6 yrs. (a) Why don't the two stars crash into one another due to the gravitational force between them? (b) What must be the mass of each star?
They don't crash into each other because they are going too fast. Gravity is not strong enough to get them any closer, and if it were to somehow get a little stronger, the stars would speed up as they move in (like an ice skater spinning and pulling in their arms) and they would reach another equilibrium point with the stronger gravity.
12.6 Earth years = (12.6 yr)(3.156x10^{7} s/yr) = 3.9766x10^{8} s
This is not a normal orbit problem, but it is not much different. The key to this problem is to realize that the planets are orbiting a point midway between them. Their radius of orbit is half their separation. Since they are 8.0x10^{10} m apart, they move in a circle 4.0x10^{10} m in radius. So now, we have gravity providing the centripetal force for them to orbit each other, so we set the force of gravity equal to the centripetal force needed for them to orbit the center. The only hitch is that if "r" is the radius of orbit, the center to center distance is now 2r. So let's set up the orbital condition with the period in it:
G(m_{1}m_{2})/(2r)^{2} = m_{1}(4π^{2}r)/T^{2 }Since both masses are the same, I will drop the subscripts:^{ }Gm^{2}/(2r)^{2} = m4π^{2}r/T^{2}_{ }^{ } Canceling m:
Gm/(2r)^{2} = 4π^{2}r/T^{2}_{ }^{ } Cross multiplying:
GmT^{2} = 4π^{2}r(2r)^{2} ^{ }=^{ }16π^{2}r^{3} Dividing both sides by GT^{2}:
m = 16π^{2}r^{3}/{GT^{2}} = 16π^{2}(4.0x10^{10} m)^{3}/{( 6.67 x 10^{11} Nm^{2}/kg^{2})(3.9766x10^{8} s)^{2}} = 9.5821x10^{26} kg = 9.6x10^{26} kg
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78. Astronomers using the Hubble Space Telescope have recently deduced the presence of an extremely massive core in the distant galaxy M87, so dense that it could well be a black hole (from which no light escapes). They did this by measuring the speed of gas clouds orbiting the core to be 780 km/s at a distance of 60 lightyears (5.7 x 10^{17} m) from the core. Deduce the mass of the core, and compare it to the mass of our Sun.
780 km/s = 7.80x10^{5} m/s
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m_{1}m_{2})/(r^{2}), G = 6.67 x 10^{11} Nm^{2}/kg^{2}
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m_{1}m_{2})/(r^{2}) = ma  where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
or
G(m_{1}m_{2})/(r^{2}) = m_{1}(4π^{2}r)/T^{2 }
But in this case, since they have given us the velocity, I am going to use the first expression:
G(m_{1}m_{2})/(r^{2}) = m_{1}v^{2}/r
Let's solve for the mass of the core (m_{2}):
G(m_{2})/(r^{2}) = v^{2}/r  (cancel satellite mass)
G(m_{2}) = v^{2}r  (Multiply both sides by (r^{2}))
m_{2} = v^{2}r/G  (Divide both sides by G)
m_{2} = v^{2}r/G = (7.80x10^{5} m/s)^{2}(5.7x10^{17} m)/(6.67 x 10^{11} Nm^{2}/kg^{2}) = 5.2x10^{39} kg or 2.6 billion solar masses (The sun has a mass of 1.99x10^{30}kg, 5.2x10^{39} kg/1.99x10^{30}kg = 2.6 billion)
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