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NASA tries to launch a 4000kg spacecraft by sending it up a 30 degree frictionless slope with an inital velocity of 200m/s. how far up the ramp does the spacecraft go before it comes to rest and does the spacecraft make it to space.
I don't know hos to handle the fact that as the higher it goes the less the pull as a result of gravity.
@radar "The higher it goes the less the pull" is the case considered if and only if the height gained by the spacecraft is comparable to earth's radius
|dw:1436675377890:dw| as shown in the figure, the component of gravitational force acting down the ramp on the spacecraft is mg/2 so it's acceleration is directed down the ramp and is g/2 in magnitude. Now, you can use the kinemetical equation v^2 - u^2 = 2as where u=200m/s, v=0m/s, a=g/2 and we have to find s. The spacecraft will not escape to space as it's initial velocity is much less than the escape velocity of the earth
@rajat97 I was just wondering because of the Newton Law where the force of gravity is equal directly to the Product of the two masses and inversely to the distance squared. The distance would be constantly increasing thus so would the decrease of the gravitational pull. I am assuming that since the mass of the earth so far exceeds the 4 million grams of the spacecraft that this can be ignored......
yes @radar you are right the distance would be constantly increasing but it will always be negligible w.r.t. the huge radius of 6400km of the earth so we can apply kinematics here
The escape velocity for earth is 11,100 meters per second. Insure that you use correct units, note that 4,000 kg is 4,000,000 grams
oh ok it makes much more sense now however i don't understand why you are dividing mg by 2.
@arturos The downward force against the frictionless ramp is equal to the the sine of 30 degrees. Sine 30 is 1/2
ah ok i see thnx!!!