The first U.S. satellite, the 14-kg Explorer 1, launched in March 1958, waS placed in an orbit for which the distances from the center of the earth ot perigee and apogee where rp= G6650 km and rA= 9920 km. (A.) Find the mechanical energy of the satellite (B.) Find the period
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- Estimate the kinetic energy of the Mars with respect to the Sun as the sum of the terms, that due to its daily rotation about its axis, and that due to its yearly revolution about the Sun. [Assume the Mars is a uniform sphere with mass = 6.4×1023 kg , radius = 3.4×106 m , rotation period 24.7 h , orbital period 686 dd and is 2.3×108 km from the Sun.] Express your answer to two significant figures and include the appropriate units.Hello, here is my question Mars has a mass of 6.417*1023 kg, a radius of 3390 km, and a rotation period of 24.62 hours. (a) Calculate the acceleration due to gravity on the surface of Mars. (b) What is the period of a simple pendulum of length 1.000 meters on the surface of Mars? (c) What is the radius of a Mars-synchronous orbit? (d) What is the speed of a satellite in a Mars-synchronous orbit? (e) What launch speed from Mars’s surface is required to get a satellite into Mars-synchronous orbit? You may neglect air resistance and any rotation effects at the Martian surface in your analysis. The universal gravitational constant is G = 6.674*10-11 N m2 kg-2 Thank you!.For a satellite orbiting the earth, the distance D from the center of the Earth is a powerfunction of the period P. Let k denote the power. So D = cP 0.669 . If the distance from the center of the Earth to one satellite is twice that of a second, how do their periods compare?
- An asteroid, whose mass is 3.6 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 1.7 times the Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth? (a) Number i (b) Number Units UnitsA Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 46,000.0 km as measured from the center of Earth and a velocity at apogee of 2.3 km/s, what would be its velocity (in km/s) at perigee measured at 540.0 km altitude? (Enter the magnitude.) 8 9. 3. 10 11 1 (time in hours) km/sA mass m = 77 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.9 m and finally a flat straight section at the same height as the center of the loop (19.9 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.) ) What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? m/s What height above the ground must the mass begin to make it around the loop-the-loop? m If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? m/s If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the…
- An asteroid, whose mass is 2.7 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 1.6 times the Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?An asteroid, whose mass is 1.9 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 3.2 times the Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?Problem 6 : Estimate the kinetic energy of the Mars with respect to the Sun as the sum of the terms, that due to its daily rotation about its axis, and that due to its yearly revolution about the Sun. [Assume the Mars is a uniform sphere with mass = 6.4×1023 kg , radius = 3.4×106 m , rotation period 24.7 h , orbital period 686 d and is 2.3×108 km from the Sun.] Express your answer to two significant figures and include the appropriate units. (Kdaily+Kyearly=?)
- A disk with velocity v= 10m/s, mass m=5 kg, and radius r=2 m. Find the total kinetic energy.Question 1 An Earth satellite has a speed 28,070 km/h when it is at its perigee of 220km above Earth's surface. Find the apogee distance, its speed at apogee, and its period of revolution. 4 Question 2A point mass m = 3 kg can move on a straight track from point A to point B. The points are on the parameter of a circular table of radius R = 8.8 m which is rotating around its center counterclockwise at constant angular velocity w = 3 rad/s. ŷR B R R At t = () the mass is at point A and begins to move at constant speed vo = 2 m/s to point B. Note that the speed remains constant during the process. What is the Coriolis force on the mass during its motion from point A to point B? a. 25.46*(Xr+Yr) O b. 50.91*(-Xr-Yr) O c. 36.00*(Xr-Yr) O d. 9.00*(-Xr+Yr) O e. 12.73*(Xr+Yr) Of. 36.00*(Xr+Yr) O g. 36.00*(-Xr+Yr) Oh. 25.46*(Xr-Yr)