Question

In a newly discovered exoplanetary system one planet orbits its star in 56 days with an average orbital radius of 107,588,122 kilometers. It is observed that another planet in the same system has orbital period of 373 days. What is its average orbital radius of the second planet? Express your answer in kilometers, use scientific notation and round the coefficient to two decimals. In a newly discovered exoplanetary system one planet orbits its star in 57 days with an average orbital radius of 99, 197,058 kilometers. It is observed that another planet in the same system has orbital distance of 726,047,548 kilometers. What is the second planet's orbital period? Express your answer days (enter"days" for units) and round it to the nearest integer. B. Schwarzschild radius (sometimes referred to as the gravitational radius) is the distance from the center of an object such that, if all the mass of the object were compressed within a sphere of such radius, the escape speed from the surface would equal the speed of light. It, thus, defines a spherical boundary called the event horizon, commonly associated with black holes, beyond which the events cannot affect an outside observer. Theoretically, any amount of matter will become a black hole if compressed into a space that fits within its corresponding Schwarzschild radius. Using equation for the escape velocity v..., 2GM and setting it equal to the speed of light vescape=[= 3 x 108 m/s, we can find the Schwarzschild radius: roman - 6 2GM Vr From this expression you can see that the Schwarzschild radius depends only on the object's mass Calculate Schwarzschild radius for the Earth (MEarth = 5.97 x 1024 kg): C. centimeters (round your answer to two decimals) Calculate Schwarzschild radius for the Sun (MSun - 1.99 % 1030 kg): D. kilometers or E. miles (round your answers to two decimals)

Answer 1

We know that m & T3 or p = KT3 A? Her r=107588122 km 7 10.76x10tum - 56 days = (56x24x360) sec = 4838400 sec = 48x 10t see Now K= m T3 E (107588122) = 1.02x10-4 (9838400 13 so să= 1002810-4 (373 X24x3600) 3 = 3.4101018 or r₂ = √3.41x10r = 1847710587 lem BY Here K= m T3 = (94, 197,658)" (57x 24 x3600) 099,197,058) – 9.24x10-5 (4924800)3 So, Tz? V = (726047548576 =6.39x1021 on, T2= (8.25 X 105) 18556328,8 see. 215 26M - 2x6.674410-145.977134 & The redives for earth is re = 20 = 3 (3x108) = 0.0088 m=0.88 cm Dy for sun 8=29M = 2x 6674x101x1.99 x1030 (38108 - 2951.39 m = 2.95km = 1.83 miles.