Suppose the space shuttle is in orbit 404 km from the Earth\'s surface, and circ

Question

Suppose the space shuttle is in orbit 404 km from the Earth's surface, and circles the Earth about once every 92.6 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth's surface. An object moves in a circle of radius 19 m with its speed given by v = 3.6 + 2.0 t^2, with v in meters per second and t in seconds. (a) At t = 2.6 s, find the tangential acceleration. m/s^2 (tangent to the path) (b) At t = 2.8 s, find the radial acceleration. m/s^2 (toward the center of the circle) Calculate the force of Earth's gravity on a spacecraft 32000 km (5 earth radii) above the Earth's surface if its mass is 1000 kg. N

Explanation / Answer

7.

time period of space shuttle, T = 92.6 min = 92.6 x 3600 sec

angular speed = 2pi/T = 2pi / (92.6 x 3600) = 1.885 x 10^-5 rad/s

and r = 6371 + 404 = 6775km = 6775000 m

Centripetal acc. = w^2 r = (1.885 x 10^-5)^2 ( 6775000) = 2.535 x 10^-3 m/s^2

in terms of g , a_c = (2.535 x 10^-3)/9.8 = 2.586 x 10^-4 g

8.

v = 3.6 + 2t^2

a) for tangential acc.

a_t = dv/dt = 0 + 4t

at t = 2.6 sec

a_t = 4 x 2.6 = 10.4 m/s^2 .............Ans

b) for radial acc.

v = 3.6 + 2t^2

first v at t = 2.8 s

v = 3.6 + 2(2.8^2) = 19.28 m/s

radial acc = v^2 / r = 19.28^2 / 19 = 19.56 m/s^2 ............Ans

9. F = G M m / r^2

F = (6.67 x 10^-11 x 1000 x 5.972 x 10^24 ) / ( (6371 + 32000) x 10^3)^2

F = 270.54 N

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