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- Problem 3: The diameter of the Sun is 1392700 km. It is known that the Sun subtends at an angle of about half a degree in the sky, estimate its distance from Earth.5. A planet of mass m is located on a line equidistant from two identical stars of mass M, located at I = +d, as shown. a) Obtain the equation of motion of the planet. b) Describe the motion (in words). M MItem 4 A laser beam is directed at the Moon, 380,000 km from Earth. The beam diverges at an angle 0 (Figure 1)of 4.3x10 5 rad Part A What diameter spot will it make on the Moon? Express your answer using two significant figures. Figure <1 of 1 d = Submit Request Answer Moon Provide Feedback Earth Laser beam
- Because of the precession of the Earth’s axis, a. there are four seasons, spring, summer, fall, and winter. b. the Earth receives more solar radiation in the summer than in the winter. c. Polaris will not be the North Star in about 12,000 years. d. the lengths of the Earth’s days and nights vary throughout the year.Suppose a Moon takes 15.8 days to orbit its host planet. Convert this orbital period into Earth years (365.25 =1 Earth year). a. 0.0431-0.0435 b. 0.0331-0.0335 c. 0.0231-0.0235 d. 0.0131-0.0135Part 3 1. The diameter of the Sun is 1,391,400 km. The diameter of the Moon is 3,474.8 km. Find the ratio, r= Dsa/Dsvan between the sizes. 2. From the point of view of an obs erver on Eanth (consider the Earth as a point-like object), during the eclipse, the Moon covers the Sun exactly. Sketch a picture to illustrate this fact. Use a nuler to get a straight line. Your drawing does not need to be in scale. 3. The Sun is 1 Astronomical Unit (AU) away from the Earth. Find the distance between the Earth and the Moon in AU's using the ratio of similar triangles. Show your work. DEM= AU. Convert this to kilometers. Use 1 AU = 149,600,000 km. DEM = km.
- It is important to have an idea about the distances between and relative sizes of celestial objects in the solar system. In Part 1 we will pretend to shrink the solar system until its center piece, the Sun, is 67.3 cm in diameter. This will represent the Sun which is 1,390,000 km in diameter. The scale of our model is thus: 67.3 cm = 4.84 x 10-5 cm km Scale 1, 390, 000 km To find the size or distance between objects in centimeters for the model, simply multiply the actual size or distance in kilometers by the scale factor above. 1. Fill in following table: Quantity Actual Distance (km) Model Distance (cm) Diameter of Sun 1,390,000 Diameter of Earth 12,760 Diameter of Moon 3,480 Distance Between Earth and Sun 1.5 x 108 Distance Between Earth and Moon 384,000 Distance to Proxima Centauri 3.97 x 101314. Why does Earth not see a lunar eclipse once a month?A. Wait . . . there is a lunar eclipse every month.B. the plane of the Moon’s orbit is tilted to the plane of Earth’s orbitC. the Sun, Earth, and Moon do not line up in this order once a monthD. Because I said so.While working with part of a research team you discover a set of exoplanets in a nearby star system. One of the planets is much closer to its mother star than the other and because of this you are able to determine the average radius of the closer planets orbit to be 37.26 x 10 to the 6 kilometers the Planet complete one orbit every 53.4 days. a) what is the mass of the star in this system?
- 3. An earth satellite moves in an elliptical orbit with a period t, eccentricity ɛ, and semimajor axis a. Show that the maximum radial velocity of the satellite is 2maɛ ((t/1-e² ).1. Consider our Sun - it is in orbit around the center of our Milky Way Galaxy. The velocity of the Sun in its orbit is about 250 km/s. The distance to the center of the galaxy is about 9.1 kpc (kiloparsecs). We can use Kepler's third law to calculate the mass of the galaxy interior to the Sun's orbit. We assume that the orbit is circular so that the semimajor axis is just the radius of the circular orbit = 9.1 kpc. First we need to calculate the number of AU's in 9.1 kpc. (Note that 1 Крс - 1000 рс - 3260 1t yrs and 1 pc - 206,265 AU.) %3D a =r =9.1kpc = (9.1kpc) 1000 pc 206,265AU] 1kpc AU Sun 1pcQUESTION 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…