Question

2 (a) What is the distance from the Sun to Earth in terms of solar radii? Earth radii?

Answer 1

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) $1.3\cdot 10^6$ Earths can fit inside the Sun

Explanation:

a)

The distance between the Sun and the Earth is 150 millions km, so

$d=150\cdot 10^6 km = 1.50\cdot 10^{11} m$

The solar radius is

$r_s = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m$

Therefore the distance Earth-Sun in solar radii is

$d_s = \frac{d}{r_s}=\frac{1.50\cdot 10^{11}}{6.96\cdot 10^8}=215.5$

The Earth radius is

$r_e = 6.37\cdot 10^6 m$

Therefore the distance Earth-Sun in Earth radii is

$d_e=\frac{d}{r_e}=\frac{1.50\cdot 10^{11}}{6.37\cdot 10^6}=23,548$

b)

The speed of the solar wind is

$v=400 km/s = 4\cdot 10^5 m/s$

The value of 1 AU (Astronomical Unit) is

$1 AU = 1.50\cdot 10^{11}m$ (distance Earth-Sun)

The distance between the Sun and Mercury is:

$d=0.4 AU \cdot 1.50\cdot 10^{11}=6.0\cdot 10^{10} m$

So the time taken by a parcel of solar wind to reach Mercury is:

$t=\frac{d}{v}=\frac{6.0\cdot 10^{10}}{4.0\cdot 10^5}=150,000 s$

Converting into days ($1d=86400 s$),

$t=\frac{150,000}{86400}=1.7 d$

The distance between the Sun and Mars is:

$d=1.5 AU \cdot 1.50\cdot 10^{11}=2.25\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Mars is:

$t=\frac{d}{v}=\frac{2.25\cdot 10^{11}}{4.0\cdot 10^5}=562,500 s$

Converting into days ($1d=86400 s$),

$t=\frac{562,500}{86400}=6.5 d$

The distance between the Sun and Jupiter is:

$d=5 AU \cdot 1.50\cdot 10^{11}=7.5\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Jupiter is:

$t=\frac{d}{v}=\frac{7.5\cdot 10^{11}}{4.0\cdot 10^5}=1.88 \cdot 10^6 s$

Converting into days ($1d=86400 s$),

$t=\frac{1.88\cdot 10^6}{86400}=21.7 d$

The distance between the Sun and Uranus is:

$d=20 AU \cdot 1.50\cdot 10^{11}=3.0\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Uranus is:

$t=\frac{d}{v}=\frac{3.0\cdot 10^{12}}{4.0\cdot 10^5}=7.5 \cdot 10^6 s$

Converting into days ($1d=86400 s$),

$t=\frac{7.5\cdot 10^6}{86400}=86.8 d$

The distance between the Sun and Neptune is:

$d=30 AU \cdot 1.50\cdot 10^{11}=4.5\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Neptune is:

$t=\frac{d}{v}=\frac{4.5\cdot 10^{12}}{4.0\cdot 10^5}=11.3 \cdot 10^6 s$

Converting into days ($1d=86400 s$),

$t=\frac{11.3\cdot 10^6}{86400}=130.2 d$

c)

As we said in part a), we have:

$r_e = 6.37\cdot 10^6 m$ (radius of the Earth)

$r_s=6.96\cdot 10^8 m$ (radius of the Sun)

So the volume of the Earth can be calculated as:

$V_e=\frac{4}{3}\pi r_e^3 = \frac{4}{3}\pi (6.37\cdot 10^6)^3=1.08\cdot 10^{21} m^3$

While the volume of the Sun is

$V_s=\frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (6.96\cdot 10^8)^3=1.41\cdot 10^{27} m^3$

Therefore, the number of Earths that could fit inside the Sun is:

$\frac{V_s}{V_e}=\frac{1.41\cdot 10^{27}}{1.08\cdot 10^{21}}=1.3\cdot 10^6$

Learn more about the Solar System:

brainly.com/question/2887352

brainly.com/question/10934170

#LearnwithBrainly