2. The mean distance between the earth and the sun is 1.5 x 1011m. The mean distance between the sun and Venus is 1.08 x 108km. What is the period of Venus around the sun in earth- days?
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- 2. The mean distance between the earth and the sun is 1.5 x 1011m. The mean distance between the sun and Venus is 1.08 x 108km. What is the period of Venus around the sun in earth-days?8. The mean distance of Earth from the Sun is 149.6 x 10° km and the mean distance of Mercury from the Sun is 57.9 x 10° km. The period of Earth's revolutions is 1 year, what is the period of Mercury's revolution? a. 0.24 year on Earth b. 0.42 year on Earth C. 1.13 year on Earth d. 1.31 year on Earth 9. The planet Delta has 2 times the gravitational field strength and 3 times the radius of Earth. How does the mass of the planet Delta compare with the mass of Earth?25. The mean distance of Earth from the Sun is 149.6 x 10° km and the mean distance of Mercury from the Sun is 57.9 x 106 km. The period of Earth's revolutions is 1 year, what is the period of Mercury's revolution? a. 0.24 year on Earth b. 0.42 year on Earth C. 1.13 year on Earth d. 1.31 year on Earth
- 8. Calculate the period of Mercury around the Sun using the Earth's period and orbital radius as a reference. (TES 1.50×10¹¹ m, MS = 5.79×10¹0 m)8. Calculate the period of Mercury around the Sun using the Earth's period and orbital radius as a reference. (TES = 1.50×10¹¹ m, TMS = 5.79×10¹0 m)3. What is the period T of a planet which radius is as twice as of Earth when it completes one revolution in 875 days? (Earth radius = 6.38 x 105)
- 1. Compute for the value of the acceleration due to gravity g of an object at an altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 106 m) 2. Scientists once hypothesized the existence of a planet called Vulcan to explain Mercury's precession. Vulcan is supposed to be between Mercury and the Sun with a solar distance equal to 2/3 of that of Mercury. What would be its supposed period? 3. What is the period T of a planet which radius is as twice as of Earth when it completes one revolution in 875 days? ( Earth radius = 6.38 x 105) %3D %3D1. The earth has a mass of 5.97219 x 1024 kilograms and a radius of 3958.8 mi. If you have a mass of 145.505 Ibs, how strong is the force between you and the moon in MKS units? 2. A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of each object. 3. What will be the radial acceleration will a satellite attain if NASA was instructed to deploy a 2000-kg satellite into our orbit 186 mi above the surface of the earth? Use the data given in number 1.1. Compute for the value of the acceleration due to gravity of an object at an altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 10^6 m) 2. Scientists once hypothesized the existence of a planet called Vulcan to explain Mercury’s precession. Vulcan is supposed to be between Mercury and the Sun with a solar distance equal to 2/3 of that of Mercury. What would be its supposed period? 3. What is the period T of a planet which radius is as twice as of Earth when it completes one revolution in 875 days? (Earth radius = 6.38 x 10^5)
- 9. Which of the following is NOT one of Kepler's Laws of Planetary Motion? A. The square of a planet's period is proportional to its distance from the sun cubed, B. The area of a planet's orbital plane is inversely proportional to its speed. C. A planet sweeps out equal area in an equal time interval. D. Planets move around the sun in elliptical orbits.2. An unknown planet was accidentally discovered by NASA. It has a mean distance of 2.15 E11 meters from the sun. Assuming it has a mass of 6.02 E24 kg, how long (in Earth Years) will it take for the said planet to revolve around the sun?33. If you are at the top of a tower of height h above the surface of the earth, show that the distance you can see along the surface of the earth is approximately s = v2Rh, where R is the radius of the earth. Hints: See figure. Show that h/R = sec 0 – 1; find two terms of the series for sec 0 = 1/ cos 0, and use s = RO. _Thus show that the distance in miles is approximately v3h/2 with h in feet. Rİ