A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatori

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Problem 4P: Venus orbits 0.72 AU from the Sun. What is that distance in kilometers? (Hint: See Problem 3.)
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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth's rotation). Calculate the radius (in km) of such an orbit based on the data for the moon in this table.  

Table 6.2 Orbital Data and Kepler's Third Law
Parent
Satellite
Average orbital radius r(km)
Period T(y)
PIT(km³ /y)
Earth
Мoon
3.84x10°
0.07481
1.01x1018
Sun
Mercury
5.79x107
0.2409
3.34x1024
Venus
1.082x10
3.35x1024
0.6150
Earth
1.496x10
1.000
3.35x1024
Mars
2.279x10*
1.881
3.35x1024
Jupiter
7.783x10*
11.86
to
29.46
3.35x10*
Saturn
1.427x10"
3.35x10*
Neptune
4.497x10°
|164.8
3.35x1024
5.90x 10"
|248.3
3.33x1024
Pluto
Jupiter
lo
4.22x10
0.00485 (1.77 d) 3.19x 10-"
Europa
6.71x10"
0.00972 (3.55 d) 3.20x10-"
Ganymede 1.07x10
0.0196 (7.16 d) 3.19x10"
Callisto
1.88×10
0.0457 (16.19 d)| 3,20x10
Transcribed Image Text:Table 6.2 Orbital Data and Kepler's Third Law Parent Satellite Average orbital radius r(km) Period T(y) PIT(km³ /y) Earth Мoon 3.84x10° 0.07481 1.01x1018 Sun Mercury 5.79x107 0.2409 3.34x1024 Venus 1.082x10 3.35x1024 0.6150 Earth 1.496x10 1.000 3.35x1024 Mars 2.279x10* 1.881 3.35x1024 Jupiter 7.783x10* 11.86 to 29.46 3.35x10* Saturn 1.427x10" 3.35x10* Neptune 4.497x10° |164.8 3.35x1024 5.90x 10" |248.3 3.33x1024 Pluto Jupiter lo 4.22x10 0.00485 (1.77 d) 3.19x 10-" Europa 6.71x10" 0.00972 (3.55 d) 3.20x10-" Ganymede 1.07x10 0.0196 (7.16 d) 3.19x10" Callisto 1.88×10 0.0457 (16.19 d)| 3,20x10
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