***NOTICE THE ANSWERS I HAVE INPUT ARE INCCORECT. Block On I -- Optiona Standard

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***NOTICE THE ANSWERS I HAVE INPUT ARE INCCORECT.

Block On I -- Optiona Standard Exerc Trip to the You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem mearth 5.9742 x 1024 kg rearth -6.3781 x 106 m mmoon -7.36 x 1022 kg moon -1.7374 x 106rm dearth to moon -3.844 x 108 m (center to center) G 6.67428 x 10-11 N-m2/kg2 1) On your first attempt you leave the surface of the earth at v-5534 m/s. How far from the center of the earth will you get? 2.07 X 10 6 Submit Your submissions: 2.07 X 10*6" x Computed value: "2.07 X 10*6 Feedback: Submitted: Thursday, November 30 at 10:10 PM 2) Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as Vmin 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon! "8.4488 x 10A6" s Submit

Explanation / Answer

1) You start off with a Kinetic energy KE and travel till it becomes zero. now this loss in kinetic energy equals gain in the potential energy

0- 0.5*m*v2 = GMm*(1/r1 - 1/r0)

Putting r0 =rearth, M=mearth, m=your mass(which gets cancelled), v=5534 m/s

we get,

0.5*55342 = 6.67428*10-11*5.9742*1024*(1/r1 - 1/(6.3781*106))

solving for r1

r1=8448692 m

2) Writing energy balance according to the hint

0.5*v2min - GM/rearth -Gmmoon/(D-rearth) = 0.5*v12 - GM/(D-rmoon) - Gmmoon/rmoon

Putting all the values and solving for v1

v1= 2377 m/s

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