January is the month in which the earth's orbit brings it the closest to the sun, while July is the month when it is the furthest away. I was trying to determine at what point in its orbit the earth moves at its quickest rate of speed.
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January is the month in which the earth's orbit brings it the closest to the sun, while July is the month when it is the furthest away. I was trying to determine at what point in its orbit the earth moves at its quickest rate of speed.
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- What is the magnitude of the Earth's centripetal acceleration? Given: Acceleration due gravity (earth surface) g = 9.8 m/s2= 9.8 N/kg Gravitational constant G = 6.67×10−11N·m2/kg2 Mass of the Sun 2.00×1030kg Mass of the Earth 5.97×1024kg Radius of the Earth 6.37×106m Radius of Earth’s Orbit 1.50×1011m Velocity: 29.870 m/sGeoEye-1 is an Earth-observation satellite that provides high-resolution images for Google. The orbital period and eccentricity of GeoEye-1 are 98.33 min and 0.001027, respectively. Determine the perigee and apogee altitudes of GeoEye-1.Suppose an asteroid named Sparty has been discovered revolving around the Sun on a circular orbit with radius rs. Calculate the period of Sparty's orbit, in years. DATA for the orbit radii: Earth re 1.50×10¹¹ m; = Sparty rs = 2.75×1011 m; A: 2.482 B: 2.805 C: 3.170 D: 3.582 E: 4.047 OF: 4.574 G: 5.168 H: 5.840 Submit Answer Tries 0/99
- A 5 kg mass is spun around in a circle of radius 2m with a period of 7 s. What equation would you use to find the centripetal force acting on the mass? Calculate the centripetal force acting on the mass. Possible Formulas that can be used to answer the question: v=(2πr)/T ac=v2/r ac=(4π2r)/T2 Fc=mac Fg=mg F=(Gm1m2)/d2 g=Gm/r2 T2=(4π2/Gm)r3 v=√(Gm)/r g=9.80m/s2 G=6.67x10-11 (N∙m2)/kg2At some point during their orbit, the location of the Earth and Moon relative to the Sun will be as shown in the figure below, with the Moon at the origin of the coordinate system, rE = (-3.84 ×: < 108) | m and rs = (-1.50 × 10¹¹1) m 7.35 x 1022 kg, the Earth has a mass 5.97 x 1024 kg, and the Sun has a mass 2.00 x 1030 kg. What is the force experienced by the Moon due to the Sun and the Earth? Express your answer in vector form. Fnet = X What is the direction of the force exerted by the Sun on the Moon? What is the direction of the force exerted by the Earth on the Moon? N r's Additional Materials Reading Tutorial Sun Submit Answer Mooni Earth 2. [0/1 Points] OSUNIPHYSI 13.2.WA.0-12.TUTORIAL. As you go above the Earth's surface, the acceleration due to its gravity will decrease. Find the height above the Earth's surface where this value will be 1/254 g. X Consider Newton's Universal Law of Gravitation for objects near the surface of the Earth. m MY NOTES m. The mass of the Moon is ASK…Titania completes one (circular) orbit of Uranus in 8.71 days. The distance from Uranus to Titania is 4.36×108m. What is the centripetal acceleration of Titania?
- The astroid belt of our solar system is located between the orbits of Mars and Jupiter. An astronomer is studying an interesting asteroid and has determined that one orbit around the Sun takes it 2.83 years. How far is the asteroid from the Sun? The following constants may be helpful. Earth's distance from the Sun = 1 AU = 1.5 x 1011 m MSun = 2.0 x 1030 kg G = 6.67 x 10-11 N m2 / kg2 1 year = 3.15 x 105 s A)1.5 AU (2.25 x 1011 m) B) 2.0 AU (3.00 x 1011 m) C) 2.5 AU (3.75 x 1011 m) D) 3.2 AU (4.80 x 1011 m) E) 4.8 AU (7.20 x 1011 m)A satellite orbits earth at a distance of 670km above the surface of the earth. The radius of the earth is RE=6357km. The mass of the earth is ME = 5.97 × 1024 kg. The gravitational constant is 10-11 Nm² G = 6.67 × kg² Determine the speed of the satellite (in km). hrThe earth orbits the sun once per year (365 days) and its average orbital radius is 1.50 x 1011m. The mass of the sun estimated from this data and Kepler’s Third “Law” is? T2=(4π2/Gmsun)r3 G=6.67 x 1011Nm2/kg2
- For this problem the average distance of the earth from the sun is about 1.5×10^8km. Assume that the earth's orbit around the sun is circular and that the sun is at the origin of your coordinate system.X 5.1.97 Question Help ▼ 5x radian per hour. The equator lies on a circle of radius approximately 3000 miles. Find the linear velocity, in miles per 12 The angular speed of a point on a planet is hour, of a point on the equator. The linear speed of a point on the equator is approximately miles per hour. (Round to the nearest whole number as needed.) Enter your answer in the answer box and then click Check Answer. All parts showing Clear All Check Answer MAR 31 tv A WA train that is travelling counterclockwise (starting at the 3 o'clock position) on a circular track with a radius of 306 feet, subtends an angle of 2 radians. a) In which quadrant is the train in at that moment? b) What is the horizontal position (measured in feet) at that moment relative to the center of the track?