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Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90 ° . What is the velocity of the rivet relative to the satellite just before striking it? (c) If its mass is 0.500 g, and it comes to rest inside the satellite, how much energy in joules is generated by the collision? (Assume the satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)

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A satellite of mass 1000 kg is in circular orbit about Earth. The radius of the orbit of the satellite is equal to two times the radius of Earth. (a) How far away is the satellite? (b) Find the kinetic, potential, and total energies of the satellite.

a. 1.3 × 10 7 m ; b. 1.56 × 10 1 0 J ; 3.12 × 10 1 0 J ; −1.56 × 10 1 0 J

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After Ceres was promoted to a dwarf planet, we now recognize the largest known asteroid to be Vesta, with a mass of 2.67 × 10 20 kg and a diameter ranging from 578 km to 458 km. Assuming that Vesta is spherical with radius 520 km, find the approximate escape velocity from its surface.

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(a) Using the data in the previous problem for the asteroid Vesta, what would be the orbital period for a space probe in a circular orbit of 10.0 km from its surface? (b) Why is this calculation marginally useful at best?

a. 6.24 × 10 3 s or about 1.7 hours. This was using the 520 km average diameter. b. Vesta is clearly not very spherical, so you would need to be above the largest dimension, nearly 580 km. More importantly, the nonspherical nature would disturb the orbit very quickly, so this calculation would not be very accurate even for one orbit.

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What is the orbital velocity of our solar system about the center of the Milky Way? Assume that the mass within a sphere of radius equal to our distance away from the center is about a 100 billion solar masses. Our distance from the center is 27,000 light years.

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(a) Using the information in the previous problem, what velocity do you need to escape the Milky Way galaxy from our present position? (b) Would you need to accelerate a spaceship to this speed relative to Earth?

a. 323 km/s; b. No, you need only the difference between the solar system’s orbital speed and escape speed, so about 323 228 = 95 km/s .

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Circular orbits in [link] for conic sections must have eccentricity zero. From this, and using Newton’s second law applied to centripetal acceleration, show that the value of α in [link] is given by α = L 2 G M m 2 where L is the angular momentum of the orbiting body. The value of α is constant and given by this expression regardless of the type of orbit.

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Show that for eccentricity equal to one in [link] for conic sections, the path is a parabola. Do this by substituting Cartesian coordinates, x and y , for the polar coordinates, r and θ , and showing that it has the general form for a parabola, x = a y 2 + b y + c .

Setting e = 1 , we have α r = 1 + cos θ α = r + r cos θ = r + x ; hence, r 2 = x 2 + y 2 = ( α x ) 2 . Expand and collect to show x = 1 −2 α y 2 + α 2 .

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Practice Key Terms 9

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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