Loose Leaf For Explorations: Introduction To Astronomy
9th Edition
ISBN: 9781260432145
Author: Thomas T Arny, Stephen E Schneider Professor
Publisher: McGraw-Hill Education
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Chapter 12, Problem 2P
To determine
The value of the gravitational force.
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Earth's daylight surface disk absorbs about 1036 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? Hint: Compare the ratio of the disk area to the spherical surface area.
I need help calculating the surface gravity of the sun.
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A rotation rate, or frequency, of 500 nHz corresponds to a rotation period of 23 days—it takes 23 days for material to follow one complete circle around the Sun. Furthermore, the frequency and period are inversely proportional to one another: if one is doubled, the other is halved. Given these facts, what is the approximate rotation period (in days) for equatorial material at the Sun's surface? (Use your answer to the previous question as the rotation rate.)
Chapter 12 Solutions
Loose Leaf For Explorations: Introduction To Astronomy
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- Step-by-step, explain how energy flows from the center of the Sun to Earth.arrow_forwardWhat is the acceleration of gravity (g) at the surface of the Sun? (See Appendix E for the Sun’s key characteristics.) How much greater is this than g at the surface of Earth? Calculate what you would weigh on the surface of the Sun. Your weight would be your Earth weight multiplied by the ratio of the acceleration of gravity on the Sun to the acceleration of gravity on Earth. (Okay, we know that the Sun does not have a solid surface to stand on and that you would be vaporized if you were at the Sun’s photosphere. Humor us for the sake of doing these calculations.)arrow_forwardQUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D = 2 AU - 3.0x1011 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where o is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? O Ps ~ 2.87 x 108 Watt/meter2 O Ps ~ 5.67 x 108 Watt/meter2 O O Ps ~ 2.87 x 10 Watt/meter2 Watt/meter² Ps ~ 5.67 x 10⁹ QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per unit area at the distance of the planet as Pp, we have Pp=Ps(Sr/D)2, where…arrow_forward
- Earth's daylight surface disk absorbs about 1047 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? (Compare ratio of disk area to spherical surface area)arrow_forwardThe Sun has a radius of 6.955×105 km and a mass of 1.989×1030 kg. What is the density of the Sun? Answer: ____ kg/m3 How does this compare to the density of Earth (5500 kg/m3)? Answer: ____ times the density of Eartharrow_forwardIf the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The volume of a sphere is (4/3)π r3. The sun is a sphere with an estimated mass of 2.00×1030 kg. What exactly is the conversion process for this?arrow_forward
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