a) Distance Earth-Sun is 215.5 solar radii and 23,548 Earth radii
b)
Mercury: 1.7 days
Mars: 6.5 days
Jupiter: 21.7 days
Uranus: 86.8 days
Neptune: 130.2 days
c) [tex]1.3\cdot 10^6[/tex] Earths can fit inside the Sun
Explanation:
a)
The distance between the Sun and the Earth is 150 millions km, so
[tex]d=150\cdot 10^6 km = 1.50\cdot 10^{11} m[/tex]
The solar radius is
[tex]r_s = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]
Therefore the distance Earth-Sun in solar radii is
[tex]d_s = \frac{d}{r_s}=\frac{1.50\cdot 10^{11}}{6.96\cdot 10^8}=215.5[/tex]
The Earth radius is
[tex]r_e = 6.37\cdot 10^6 m[/tex]
Therefore the distance Earth-Sun in Earth radii is
[tex]d_e=\frac{d}{r_e}=\frac{1.50\cdot 10^{11}}{6.37\cdot 10^6}=23,548[/tex]
b)
The speed of the solar wind is
[tex]v=400 km/s = 4\cdot 10^5 m/s[/tex]
The value of 1 AU (Astronomical Unit) is
[tex]1 AU = 1.50\cdot 10^{11}m[/tex] (distance Earth-Sun)
The distance between the Sun and Mercury is:
[tex]d=0.4 AU \cdot 1.50\cdot 10^{11}=6.0\cdot 10^{10} m[/tex]
So the time taken by a parcel of solar wind to reach Mercury is:
[tex]t=\frac{d}{v}=\frac{6.0\cdot 10^{10}}{4.0\cdot 10^5}=150,000 s[/tex]
Converting into days ([tex]1d=86400 s[/tex]),
[tex]t=\frac{150,000}{86400}=1.7 d[/tex]
The distance between the Sun and Mars is:
[tex]d=1.5 AU \cdot 1.50\cdot 10^{11}=2.25\cdot 10^{11} m[/tex]
So the time taken by a parcel of solar wind to reach Mars is:
[tex]t=\frac{d}{v}=\frac{2.25\cdot 10^{11}}{4.0\cdot 10^5}=562,500 s[/tex]
Converting into days ([tex]1d=86400 s[/tex]),
[tex]t=\frac{562,500}{86400}=6.5 d[/tex]
The distance between the Sun and Jupiter is:
[tex]d=5 AU \cdot 1.50\cdot 10^{11}=7.5\cdot 10^{11} m[/tex]
So the time taken by a parcel of solar wind to reach Jupiter is:
[tex]t=\frac{d}{v}=\frac{7.5\cdot 10^{11}}{4.0\cdot 10^5}=1.88 \cdot 10^6 s[/tex]
Converting into days ([tex]1d=86400 s[/tex]),
[tex]t=\frac{1.88\cdot 10^6}{86400}=21.7 d[/tex]
The distance between the Sun and Uranus is:
[tex]d=20 AU \cdot 1.50\cdot 10^{11}=3.0\cdot 10^{12} m[/tex]
So the time taken by a parcel of solar wind to reach Uranus is:
[tex]t=\frac{d}{v}=\frac{3.0\cdot 10^{12}}{4.0\cdot 10^5}=7.5 \cdot 10^6 s[/tex]
Converting into days ([tex]1d=86400 s[/tex]),
[tex]t=\frac{7.5\cdot 10^6}{86400}=86.8 d[/tex]
The distance between the Sun and Neptune is:
[tex]d=30 AU \cdot 1.50\cdot 10^{11}=4.5\cdot 10^{12} m[/tex]
So the time taken by a parcel of solar wind to reach Neptune is:
[tex]t=\frac{d}{v}=\frac{4.5\cdot 10^{12}}{4.0\cdot 10^5}=11.3 \cdot 10^6 s[/tex]
Converting into days ([tex]1d=86400 s[/tex]),
[tex]t=\frac{11.3\cdot 10^6}{86400}=130.2 d[/tex]
c)
As we said in part a), we have:
[tex]r_e = 6.37\cdot 10^6 m[/tex] (radius of the Earth)
[tex]r_s=6.96\cdot 10^8 m[/tex] (radius of the Sun)
So the volume of the Earth can be calculated as:
[tex]V_e=\frac{4}{3}\pi r_e^3 = \frac{4}{3}\pi (6.37\cdot 10^6)^3=1.08\cdot 10^{21} m^3[/tex]
While the volume of the Sun is
[tex]V_s=\frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (6.96\cdot 10^8)^3=1.41\cdot 10^{27} m^3[/tex]
Therefore, the number of Earths that could fit inside the Sun is:
[tex]\frac{V_s}{V_e}=\frac{1.41\cdot 10^{27}}{1.08\cdot 10^{21}}=1.3\cdot 10^6[/tex]
Learn more about the Solar System:
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