Early test flights for the space shuttle used a \"glider\" (mass of 980 kg inclu

Question

Early test flights for the space shuttle used a "glider" (mass of 980 kg including pilot). After a horizontal launch at 600 km/h from a height of 3200 m, the glider eventually landed at a speed of 210 km/h. What would its landing speed have been in the absence of air resistance? Express your answer using two significant figures. What was the average force of air resistance exerted on it if it came in at a constant glide of 12 degree to the Earth? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

From conservation of energy

KE + PE = W

1/2 m v2^2 - 1/2 m v1^2 + -G M m /r2 - (- G M m/r1) = W

r = 0, at the earth's center, r1 = R + h = 6378000 + 3200 = 6381500 m ; r2 = 6378000 m

when there is no air resistance, W = 0

1/2 m v2^2 - 1/2 m v1^2 + -G M m /r2 - (- G M m/r1) = 0

m gets cancelled both sides

v1 = 600 km/h = 166.67 m/s

v2 = 210 km/h = 58.33 m/s

1/2 v2^2 - 1/2 x 166.67^2 + (-6.67 x 10^-11 x 5.97 x 10^24/6378000) - (-6.67 x 10^-11 x 5.97 x 10^24/63815000) = 0

1/2 v^2 - 13889.44 -6.24 x 10^7 + 6.24 x 10^6 = 0

1/2 v2^2 = 5.62 x 10^7 => v2 = 1.06 x 10^4 m/s

Hence, v2 = 1.06 x 10^4 m/s

b)The work done by air resistance is:

1/2 m v2^2 - 1/2 m v1^2 + -G M m /r2 - (- G M m/r1) = W

1/2 x 980 (58.33^2 - 166.67^2) + (-6.67 x 10^-11 x 5.97 x 10^24/6378000) - (-6.67 x 10^-11 x 5.97 x 10^24/63815000) = W

W = -1.2 x 10^7 - 6.24 x 10^7 + 6.24 x 10^6 = -6.81 x 10^7 J

W = -F h/sin(theta)

F = W x sin(theta)/h

F = 6.81 x 10^7 x sin12/3200 = 4424.62 N

Hence, F = 4424.62 N

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