You are on a space station, in a circular orbit h = 475 km above the surface of the Earth. You complete your tasks several days early and must wait for the next mission from the surface to bring you home. After days of boredom, you decide to play some golf. Walking on the space station surface with magnetic shoes, you tee up a golf ball. You hit it with all of your might, sending it off with speed vrel relative to the space station, in a direction parallel to the velocity vector of the space station at the moment the ball is hit. You notice that you then orbit the Earth exactly n = 2.00 times and you reach up and catch the golf ball as it returns to the space station. With what speed vrel (in m/s) was the golf ball hit? Note: Your result will be unrealistically high-much higher than it is possible for a human to hit a golf ball. m/s

When an object is orbit around a planet, what is it actually going on? The force of Earth's gravity pushes outward and keeps the object from falling. Gravity, combined with the satellite's momentum, cause the satellite to stay orbit above Earth, instead of falling back down to the ground. These forces must be in balance to maintain an orbit. The object starts to have no mass, and therefore can stay in orbit. When the object is in orbit, gravity no longer acts upon that object, and therefore can maintain its constant circular path without a influence of an outside force.

We have a non-rotating space station in the shape of a long thin uniform rod of mass 4.51 x 10^6 kg and length 1336 meters. Small probes of mass 9720 kg are periodically launched in pairs from two points on the rod-shaped part of the station as shown, launching at a speed of 4220 m/s with respect to the launch points, which are each located 387 m from the center of the rod. After 20 pairs of probes have launched, how fast will the station be spinning?     2.17 rpm     9.04 rpm     3.62 rpm     13.56 rpm

a) We send a probe to orbit a nearby asteroid and take some pictures of it. The probe enters an orbit that puts it 850m from the centre of the asteroid. If the probe moves at 12m/s, determine the mass of the asteroid

Earth's orbital velocity is about 29 km/s and that maintains its nearly circular orbit.  Suppose we had a spacecraft following Earth in its orbit, and we applied  a rocket to slow it to 25 km/s.  Assume the thrust is entirely in the plane of the orbit, so the spacecraft stays in the same plane as Earth's orbit.  What happens?     It falls in to sun and is then flung out of the solar system altogether by the energy it picks up on this path.     The spacecraft goes into a new orbit that is an ellipse with perihelion (nearest  to Sun) at Earth's orbit.     It stays in the same orbit with the Earth but takes longer to complete its new nearly circular orbit.     The spacecraft goes into a new orbit that is an ellipse with aphelion (farthest from Sun) at Earth's orbit.

A meteor of mass m is approaching earth as shown on the sketch. The distance h on the sketch below is called the impact parameter. The radius of the earth is Re 6400km . The 6x1024 kg mass of the earth is me = Suppose the meteor has an initial speed of vo = 30km/s. Assume that the meteor started very far away from the earth. Suppose the meteor just passes earth at a distance of 2.5 Re from the earth's center. You may ignore all other gravitational forces except the earth. Find the moment arm h in km (called the impact parameter). G -11 6.673x10-"Nm²kg %3D meteor very far away Vo h impact parameter planet

A binary-star system contains a visible star and a black hole moving around their center of mass in circular orbits with radii r1 and r2 , respectively. The visible star has an orbital speed of v=5.36x105 ms-1 and a mass of m1 =5Ms ,where Ms= 1.98x1030kg is the mass of our Sun. Moreover, the orbital period of the visible star is T = 30 hours.(a) What is the radius r1 of the orbit of the visible star?(b) Calculate the mass m2 of the black hole in terms of MS . [Hint: One root of the equation x3 = 20a(5a+x)2  , where a is a constant, is x = 28a .]