A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s. (a) How fast is he going as he lands on the trampoline, 3.0 m below (Fig. 6–38)? (b) If the trampoline behaves like a spring with spring stiffness constant 6.2 × 104N/m, how far does he depress it?

FIGURE 6–38

Solution Step 1 of 5 The quantities like orbital period, speed and acceleration of the space shuttle revolving around the earth, does not depend on the mass of the space shuttle. It depends on the mass of the central body around which the shuttle is revolving and the radius of the orbit. So we can calculate the above physical quantities using the given information; height of the orbit from earth surface. Orbital time period The time period in the given orbit is the time taken by the space shuttle to complete one full revolution around earth. The relation between the orbital time period T and the distance of space shuttle r from the center of the earth, 2 23 T = GM r Where T is the orbital time period, r is the radius of the orbit, G is gravitational constant and M is the mass of the earth. From the above equation, the orbital time period T is, 3/2 T = 2r GM Step 2 of 5 Space shuttle is revolving at 250 miles from the earth surface, so the total distance from the earth center will be the sum of the radius of earth and the distance of space shuttle from the surface. That is, r = R +h e Where r is the distance of shuttle from the earth center, R isethe radius of the earth and h is the height at which the shuttle is revolving around the earth. Using this in above equation, 2(Re+h)/2 T = GM Given data Height, h = 250 miles To convert height from miles to meter Using 1 mile = 1609 m h = 250 (1609)=0.40225×10 m 6 6 Radius of the earth, R =6e37×10 m 24 Mass of the earth, M=5.98×10 kg 11 2 2 G=6.67 ×10 Nm /kg To find, Orbital time period, T= Step 3 of 5 Substituting above given data in time period equation, 3/2 2(3.14)[(6.37×10 m)+(0.40225×10 m)] T = (6.67 ×101Nm /kg )(5.98×10 kg) 3 T = 5.5445×10 s Using 1 min = 60 sec T = 5.5445×10 s( 3 1 mi) 60 s T = 92.41 min Therefore, the orbital time period of the space shuttle around the earth is 92.41 min.