Question

1.The space shuttle is in a 250-km-high circular orbit. It needs to reach a 610-km-high circular orbit to catch the Hubble Space Telescope for repairs. The shuttle

Answer 1

Uf = E + Ui

(-6.67 * 10-11)(5.98 * 10²⁴kg)(75000kg)/(610000 m + 6.37* 10⁶) = E + (-6.67 * 10--11)(5.98 * 10²⁴)(75000kg)/ (250000 m + 6.37 * 10⁶)

E = 1.8 * 10¹³ J

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Potential Energy = mgh = m * 9.81 * 0.5 = 4.91 m

v=(2GM/R)^(1/2)

Kinetic Energy = 0.5 * m v² = 3.34 m

yes, you can escape from the asteroid.

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3)F_centripetal = F_{gravitational}

m_s * a_g = m_s * a_c

a_g = a_c

a_c = (omega)² * r, a_g = G M / r²

(omega)² * r = G M / r²

r = cubic root [GM / (omega)²]

speed of a geosynchronous satellite = cubic root (GM*omega) = cubic root (G M * 2 pi / T)

= cubic root [(6.673 x 10-11 x 6.4191 x 10²³ kg x 2 x pi) / (24.8hours x 3600 seconds/hr)]

= 1440 m/s = 1.44 km/s

v = omega * r, r = v / omega

omega = 2pi / T = 2 * 3.14 / (24.8hours x 3600 seconds/hr)]

v = 1440 m/s

r = 2.05 * 10⁷

radius of mars = 3397000 m

Altitude = 2.05 * 10⁷ - 3397000 = 1.72 * 10⁷m

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Difference in potential energy = -G2M^2 ( 1/10R - 1/2R)

= 4 G M²/ 5 R

By conservation of energy

4 G M²/ 5 R = 1/2 M v1²+ 1/2 2 M v2²

v1²+2 v2²= 8 G M/ 5 R

Also momentum is conserved M v1 = 2 M v2

v1 = 2 v2

6 v2²= 8 G M / 5 R

v2 = sqrt (4GM/15R) = 2 sqrt (GM/15R)

v1 = sqrt( 16GM/15R) = 4 sqrt (GM/15R)