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# gp5 - Physics 106 Spring 2008 K Kauder Group Problem#5...

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Physics 106, Spring 2008 K. Kauder Group Problem #5 –Answer key– 1. Gravity A 2000 kg spaceship is initially traveling in a circular orbit around a planet at a distance four times the radius of the planet as measured from the planet’s center. The radius of the planet, R p is 400 km. The spaceship goes into a new circular orbit whose radius is sixteen times the radius of the planet as measured from the planet’s center. The planet’s mass is unknown but the acceleration due to gravity on its surface g p has been measured to be 2.5 m s 2 . (a) Determine the planet’s mass M knowing the value of the acceleration due to gravity on its surface. (b) Determine the speed of the spaceship when it is in its initial orbit. (c) Determine the work done by the spaceship’s en- gine in traveling from the initial orbit to the final one. Solution : (a) We can express the force acting on some test object of mass m on the surface in two ways: universal law of gravitation: F = G m M R 2 p and local law of gravitation: F = m g p These are two ways to describe the same force F so we can equate the two terms: G m M R 2 p = m g p Conveniently, the test mass m can be eliminated and we can rearrange the expression to find the solution: M = R 2 p g p G = (400 × 10 3 m) 2 (2 . 5 m s 2 ) 6 . 67 × 10 - 11 N m 2 kg 2 = 6 . 0 × 10 21 kg (b) Reminder: Paraphrasing the textbook (about centripetal force , equation 6–15 or thereabouts): “A force must be applied to an object [with tangential speed v] to give it circular motion. For an object of mass m the net force acting on it must have a magnitude given by f cp = m a cp = m v 2 r and must be directed toward the center of the circle.” Luckily, we are concerned with circular motion (treating a general elliptic orbit would be a bit nastier). Here, the centripetal force is supplied by gravity so we can equate the two and solve for v (note that r = 4 R p ): G m M r 2 = m v 2 r v 2 = G M r v = radicalBig G M r = radicalBig 6 . 67 × 10 11 N m 2 kg 2 6 . 0 × 10 21 kg 4(400 × 10 3 m) = 500 m s

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Physics 106/GP5 – Page 2 of 2 – –Answer key– (c) In order to determine the amount of work done by the rocket’s engine we must write down the most general form of the work-energy theorem which is W = Δ K + Δ U where Δ U is the change in the gravitational potential energy of the astronaut. We can write this equation explicitly as
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gp5 - Physics 106 Spring 2008 K Kauder Group Problem#5...

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