webct10 - Spacecraft Ten days after it was launched toward...

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Spacecraft Ten days after it was launched toward Mars, a spacecraft with mass 629 kg is 2.87 x 10 6 km from the Earth and traveling at 1.25x 10 4 km/h relative to the Earth. 1. At this time, what is the spacecraft's kinetic energy relative to the Earth, in J? The kinetic energy is 2 1 2 Km v = We need to convert the speed to m/s: 4 km 1000 m 1 h 1.25 10 3472 m/s h 1 km 3600 s v = () 2 29 11 (629 kg) 3472 m/s 3.79 10 J 22 v == = × 2. And what is the gravitational potential energy of the spacecraft, in J? (Take the gravitational potential energy to be zero at infinity). When the potential energy is zero at infinity, the potential energy function is ( )( ) 24 2 11 7 5.98 10 kg 629 kg Nm 6.67 10 8.74 10 J C2 . 8 7 1 0 m Mm UG r × =− × × × (mass of the Earth M = 5.98x10 24 kg)
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Comet 1. The only force that keeps planets (or other objects, like comets) in orbit around the Sun is Newton's gravitational force ˆ 2 Mm F Gr r =− G . The torque produced by this force about the Sun: *a. Is zero. b. Points towards the Sun. c. Points away from the Sun. The angle between the positon vector r & force vector F is 0 o (they are parallel). Hence the torque produced by this force about the Sun is , sin 0 0 o rF τ = = 2. As a consequence, the angular momentum of the planet/comet: a. Increases with r , the distance to the Sun. b. Decreases with r , the distance to the Sun. *c. Is constant along the orbit (conserved). The torque is zero and hence the angular momentum of the planet conserved, ie remains constant along the orbit. 3. The orbit of many comets, like Halley's (shown in the figure below), is an ellipse, the Sun being at one focus. Consider a comet with an elliptic orbit whose maximum distance to the Sun (aphelion distance) is r a = 5.00 × 10 9 km. The minimum distance
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(perihelion distance) is r p = 8.00 × 10 7 km. Find the speed of this comet at the perihelion position. Angular momentum is conserved, aphelion perihelion aa pp (1) LL mv r mv r vr = = = But we have two unknowns in one equation. The other condition that must be respected is conservation of mechanical energy: aphelion perihelion 22 Sun Sun Sun 11 2 (2) ap EE Mm mv G mv G rr vv G M = −=    (1) and (2) make a system of two equations with two unknowns. Solving for v a in (1) and substituting it in (2), we obtain:
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2 2 Sun 2 11 12 p p aa p r vG M rr r  −=   Thus, () Sun 2 2 Sun 2 22 Sun 2 Sun 2 1 1 ap p p a p p a pa p M r r GM r GM r r GM rr r = = = + = +
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This homework help was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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webct10 - Spacecraft Ten days after it was launched toward...

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