At the distance of Earth, 1AU, the Sun's apparent brightness in the sky is about 1300...
At the distance of Earth, 1AU, the Sun's apparent brightness in the sky is about 1300 W/m2. What is the brightness that we would measure from a distance of 0.5 AU?
Answer:
11. [3pt]
What is the brightness that we would measure from a distance of 2.0
AU?
Answer:
12. [3pt]
What is the brightness that we would measure from a distance of 15
AU?
Answer:
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At the distance of Earth, 1AU, the Sun's apparent brightness in
the sky is about 1300...
. [5 marks] The average apparent brightness of a particular Type I Cepheid variable in a...
. [5 marks] The average apparent brightness of a particular Type I Cepheid variable in a distant galaxy is found to be 4.30 x10-17 W/m2 ; the period of the Cepheid is 30 days. What is the distance to the galaxy? Give your answer in parsecs and in light years. [Make use of Figure 19-20 on page 553 of our textbook to read off what the luminosity of this Cepheid would be in units of L; use the vertical scale...
The apparent brightness of a star is 1.0 × 10-12 W/m2 and the peak wavelength is...
The apparent brightness of a star is 1.0 × 10-12 W/m2 and the peak wavelength is 600 nm. Estimate its distance from us. 0.090 pc 0.90 pc 900 pc 9.0 pc 90 pc I do not understand how 179.24 pc relate to the answer choices I am given.
Consider a bright star in our night sky. Assume its distance from Earth is 89.1 light-years...
Consider a bright star in our night sky. Assume its distance from Earth is 89.1 light-years (ly) and its power output is 4.00 ✕ 1028 W, about 100 times that of the Sun. One light-year is the distance traveled by light through a vacuum in one year. (a) Find the intensity of the starlight at the Earth. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. W/m2 (b) Find the...
Warm Up The surface of the Earth receives approximately 1000 W/m2 (Watts per square meter) of energy from the Sun....
Warm Up The surface of the Earth receives approximately 1000 W/m2 (Watts per square meter) of energy from the Sun. This number rises to about 1386 W/m2 in the upper atmosphere (r 6.4 x 106 m) The sun is 1 AU (1.496 x 1011 m) away Over what area is the Sun's energy spread at the Earth's distance? How much power is the Sun releasing?
Warm Up The surface of the Earth receives approximately 1000 W/m2 (Watts per square meter)...
Sun's rays at noon 496 mi An estimate of the distance around the earth was done...
Sun's rays at noon 496 mi An estimate of the distance around the earth was done by a scientist who noted that the noontime position of the sun differed by 7°12' from one city to the next closest city. (See the figure.) The distance between these two cities is 496 mi. Use the arc length formula to estimate the radius of Earth. Then find the circumference of Earth. 7° 12 Shadow City A City B 7° 12 The radius of...
The intensity of sunlight hitting the earth is about 1300 W/m2. If sunlight strikes a perfect...
The intensity of sunlight hitting the earth is about 1300 W/m2. If sunlight strikes a perfect absorber, what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to?
Consider a bright star in our night sky. Assume its distance from Earth is 31.7 light-years...
Consider a bright star in our night sky. Assume its distance from Earth is 31.7 light-years (ly) and its power output is 4.00 x 1028 W, about 100 times that of the Sun. (a) Find the intensity of the starlight at the Earth. 3.54e-8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nW/m² (b) Find the power of the starlight the Earth intercepts. One light-year is the distance traveled...
Question 7 (0.5 points) What causes the Moon to move about 12° across the sky from...
Question 7 (0.5 points) What causes the Moon to move about 12° across the sky from one night to the next (at the same time of night, of course)? O Because the Earth is turning on its axis. O Because the Moon is moving in its orbit. The Sun has also moved 15° across the sky and gravitationally pulls the Moon with it. The celestial sphere the Moon is attached to has moved 15°. O It is an optical illusion....
9) How distant is a star whose modulus is -7.5? 10) If its apparent magnitude (m)...
9) How distant is a star whose modulus is -7.5? 10) If its apparent magnitude (m) = +13.5 and absolute magnitude (M) = +10, what is the distance to a star? PROBLEMS TO COMPLETE FOR PARALLAX 11) A star has a parallax of .025 seconds of arc—what is its distance in parsecs? The nearest star has a parallax of 76 arc seconds—what is its distance in light years? 12) If star were 2.40 pc from Earth, what would its distance...
2) Planet Velocities and Energy (10 pts) We talked about how planet formation involves the collisions...
2) Planet Velocities and Energy (10 pts) We talked about how planet formation involves the collisions of bodies (planetesimals, embryos) leading to the growth (and heating) of a planet. Let's think about the velocities and energies involved here. a) The speed of a body in its orbit around the Sun is given by the equation V2= GM.[(2/r) - (1/a)] Here Vis the speed of the body in m/s, G is the gravitational constant, M. is the mass of the Sun...
Warm Up The surface of the Earth receives approximately 1000 W/m2 (Watts per square meter) of energy from the Sun. This number rises to about 1386 W/m2 in the upper atmosphere (r 6.4 x 106 m) The sun is 1 AU (1.496 x 1011 m) away Over what area is the Sun's energy spread at the Earth's distance? How much power is the Sun releasing?
Warm Up The surface of the Earth receives approximately 1000 W/m2 (Watts per square meter)...
Sun's rays at noon 496 mi An estimate of the distance around the earth was done by a scientist who noted that the noontime position of the sun differed by 7°12' from one city to the next closest city. (See the figure.) The distance between these two cities is 496 mi. Use the arc length formula to estimate the radius of Earth. Then find the circumference of Earth. 7° 12 Shadow City A City B 7° 12 The radius of...
Consider a bright star in our night sky. Assume its distance from Earth is 31.7 light-years (ly) and its power output is 4.00 x 1028 W, about 100 times that of the Sun. (a) Find the intensity of the starlight at the Earth. 3.54e-8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nW/m² (b) Find the power of the starlight the Earth intercepts. One light-year is the distance traveled...
Question 7 (0.5 points) What causes the Moon to move about 12° across the sky from one night to the next (at the same time of night, of course)? O Because the Earth is turning on its axis. O Because the Moon is moving in its orbit. The Sun has also moved 15° across the sky and gravitationally pulls the Moon with it. The celestial sphere the Moon is attached to has moved 15°. O It is an optical illusion....
9) How distant is a star whose modulus is -7.5? 10) If its apparent magnitude (m) = +13.5 and absolute magnitude (M) = +10, what is the distance to a star? PROBLEMS TO COMPLETE FOR PARALLAX 11) A star has a parallax of .025 seconds of arc—what is its distance in parsecs? The nearest star has a parallax of 76 arc seconds—what is its distance in light years? 12) If star were 2.40 pc from Earth, what would its distance...
2) Planet Velocities and Energy (10 pts) We talked about how planet formation involves the collisions of bodies (planetesimals, embryos) leading to the growth (and heating) of a planet. Let's think about the velocities and energies involved here. a) The speed of a body in its orbit around the Sun is given by the equation V2= GM.[(2/r) - (1/a)] Here Vis the speed of the body in m/s, G is the gravitational constant, M. is the mass of the Sun...
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