Planet A is located 1.81 x 1010 [m] away from its parent star. What is the radial acceleration of Planet A if it revolves around its parent star at a constant angular speed of 3.97 x 10-6 [rad/s] along a circular orbit?
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- Planet A is located 1.81 x 10^10 [m] away from its parent star. What is the radial acceleration of Planet A if it revolves around its parent star at a constant angular speed of 3.97 x 10-6 [rad/s] along a circular orbit?An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 1810 × 103 seconds (about 21 days) on average to complete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 255.0 × 106 m and the planet has a radius of 3.30 × 106 m, calculate the moon's radial acceleration ?cac.Take the radius of the Earth to be 6309 km. The angular speed of a point on Earth’s surface at latitude 30° N is Blank 1 x 10-4 rad/s. The linear speed of a point on Earth’s surface at latitude 30° N is Blank 2. The latitude where your linear speed is 10 m/s is ±Blank 3°. Round your answer to the nearest 2 decimal places.
- 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 WTwo celestial bodies whose masses are m1 and m2 are revolving around their common center of mass and the distance between them is L. Assuming that they are both point masses, Find the angular speed, tangential speeds of the masses m1 and m2, and period of the motion. Universal Gravitational Constant, G=6,6742867E-11 m3 kg / s2(Note that the exponent is negative)Radius of Earth, RE: 6,3781366E+06 mMass of Earth, ME: 5,9721426E+24 kg m1=10^12kg m2=10^11kg L=10^8m 7,27210E+00 m1 3,85280E+00 m2 6,16500E+00 LI need some help solving this problem: The angular speed of a point on a planet is 5π/11 radian per hour. The equator lies on a circle of radius approximately 6000 miles. Find the linear velocity, in miles per hour, of a point on the equator. Round to the nearest whole number as needed. Thanks so much for your help!
- What is the tangential velocity (in kilometers per hour) of the earth at the equator? The radius of the earth is 6,371 km.a) What is the period of rotation of Venus in seconds? (The period of rotation of Venus in hours is 5,832.5 hr.) Answer in seconds (b)What is the angular velocity (in rad/s) of Venus? (Enter the magnitude.) Answer in rad/s (c) Given that Venus has a radius of 6.1 ✕ 106 m at its equator, what is the linear velocity (in m/s) at Venus's surface? (Enter the magnitude of the linear velocity at the equator.) Answer on m/sQuestion 1 a. Suppose the Sun contracts (collapses) to a pulsar with radius 15 km. Estimate the period of rotation of the pulsar. The period of revolution of the Sun about its axis is 25.38 days (1 day = 24 hours). Compare the kinetic energies of rotation of the pulsar and of the Sun. Figure 1: Question 1(a) A pulsar star system b. A wheel on an extremely low friction slanted surface will slide without rotating. However, if the same surface is rough and having some friction the wheel will roll down (with rotation). Explain why, using free body diagram.
- A star with a of mass of 3.0x1032 kg and radius 7.0x108 m is initially rotating at a rate of once every 30 days. The star collapses into a neutron star with the same mass but a new radius of 18,000 m. What is the new angular speed of the star? (Give your answer in rotations per second.) Assume the star is a solid sphere: Isphere = 2/5 MR2.The Earth is 93 million miles (mi) from the Sun and its period of revolution is 1 year = 31,500,000 s. What is the radial acceleration of the Earth in its orbit about the Sun? Choose 0.0093 mi/s^2 18.6 mi/s^2 0.0000037 mi/s^2 0.0000136 mi/s^2A spaceship orbits earth from a distance of 2.656 × 10^4km with a centripetal acceleration of 7.962 × 10^3 m/s, if the astronauts mission time period is 16 months, how many rotations will the ship complete before returning to earth? ( earth's circumference= 4.0075 × 10^4km)