The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of 3.0 x 10 m/s. Astronomers have determined that the solar system is orbiting the center of the galaxy at a speed of 230 km/s. Assume that the sun is a typical star with a typical mass. If galactic matter is made up of stars, approximately how many stars are in the center of the galaxy? Note: Astronomers have spent many years trying to determine how many stars there are in the Milky Way. The number of stars seems to be only about 10% of what you'll find in part d. In other words, about 90% of the mass of the galaxy appears to be in some form other than stars. This is called the dark matter of the universe. No one knows what the dark matter is. This is one of the outstanding scientific questions of our day. 15. ΑΣΦ N=6.810¹2 Submit Previous Answers Request Answer X Incorrect; Try Again ? stars

The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of 3.0 x 10 m/s. Astronomers have determined that the solar system is orbiting the center of the galaxy at a speed of 230 km/s. Assume that the sun is a typical star with a typical mass. If galactic matter is made up of stars, approximately how many stars are in the center of the galaxy? Note: Astronomers have spent many years trying to determine how many stars there are in the Milky Way. The number of stars seems to be only about 10% of what you'll find in part d. In other words, about 90% of the mass of the galaxy appears to be in some form other than stars. This is called the dark matter of the universe. No one knows what the dark matter is. This is one of the outstanding scientific questions of our day. 15. ΑΣΦ N=6.810¹2 Submit Previous Answers Request Answer X Incorrect; Try Again ? stars

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A spacecraft is traveling at ~v = 38.4km s (yes, that is a reasonable speed for a rocket). How long will it take to travel from the earth to the moon, a distance of 384 Mm from Earth?
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A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 48,300 ly (light-years) from the galactic center and has a speed of 275 km/s. Estimate the total mass of the galaxy based on the motion of the star. Gravitational constant is 6.674×10−11 m3/(kg·s2) and mass of the Sun Ms=1.99 × 1030 kg. dont provide hand written solution
Sam is an astronomer on planet Hua, which orbits the distant star Barnard. It has recently been accepted that Hua is spherical in shape, although its exact size is unknown. While studying in the library, in the city of Joy, Sam learns that during equinox, Barnard is directly overhead in the city of Bar, located 1500.0 km north of his location. On the equinox, Sam goes outside and measures the altitude of Barnard at 83 degrees. What is the radius of Hua in km?
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How long should it take the voices of astronauts on the Moon to reach the Earth? Assume that the only significant time is the transit time from the Earth to the Moon, at the speed of light. Suppose that the astronaut on the surface of the Moon, the receiver on the surface of the Earth, and the centers of the Earth and the Moon are aligned. The distance between the centers of the Earth and the Moon is 384×103km, the radius of the Earth is 6.38×103km, the radius of the Moon is 1.74×103km.
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The class I'm taking is physics for scientists and engineers! I am completely stuck. Need help. I have attached the problem. Please view both attachments before answering. Please write step-by-step solution so I can fully understand.
We will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→ f=Gmm/lr^→l^2 a^→=Gmm/lr^→l^2 x r^→/lr^→l r^→=r^→/lr^→l * Gmm Could you please…
A sun rays take about 8 minutes to reach the Earth surface travelling at constant speed v= 3,0 * 10^8 m/s. How far is earth from the sun?