Normally, the potential energy of a planet at a given distance is negative- it is taken so that the potential energy "at infinity", at an infinite distance from the planet is 0.
The escape velocity is the velocity at which the object has sufficient kinetic energy so that its total energy (kinetic plus potential) is 0. That way, without any other force (i.e. not a rocket) it will go as far from the planet as you please.
I don't understand the "constant speed" part. If you can set up the forces (rocket or whatever) so that the object moves with constant velocity, then it will escape eventually, no matter how low that velocity is!
The escaping process takes place in two steps, we apply Conservation of energy at the two extreme ends of this complex process.
During the launch of the object , it is believed to have the potential energy due to gravitation field of the planet and the needed kinetic energy , then after it takes off , and eneters the space , it is assumed to travel far off and then sail in space (with negligible velocity) where the potential energy is taken to be zero.From here we calculate the escape velocity .
So it is basically the process in which earth's potential energy is being lost to gain the required 'v' which is the escape velocity , to overcome the earth's gravitational pull and sail off.
SORRY PPL, this is not a homework question, its a high level provincial june 2005 question, i was just wondering if i got it correct, by the way i did pick option number 3 but i did waste 30 mins doing irrelevent calc to find out how to get CONSTANT speeed
The escape velocity is irrelevent, I think. Since the object is required to have a constant speed, its kinetic energy is the same in the end as it is in the beginning. The only relevant change is the change in potential energy, I think that's all there is to the problem.
easy to say 5 mins, but how abt u actually think abt this question!! it says COnstant speed, that mean if a rocket has 100 m/s then it continues at that speed yet it escapes the gravitational field of a planet. DO U GUYZ THINK THIS IS A RELEVENT QUESTION FOR PROVINCIAL??? I DON'T, CAZ THE CONSTANT SPEED THING IS BULLCRAP.
The relevance of constant speed is that it means the change in kinetic energy is zero. Suppose the object was moving at escape velocity to begin with. If the problem did not say the object moved with constsant speed you would not have to add any energy to get the object to infinity, but the speed of the object there would be zero. Adding the requirement of constant speed means the object is still moving with escape velocity at infinity, so that you still must add an amount of energy equal to the change in potential energy. If the object is initially moving at less than escape velocity, the amount of energy you need to add is still the same, because it will end moving at the same speed it started with.
Refer to LeonhardEuler's explanation (#7) , the answer is worth 5 seconds , should not take 5 minutes.
Oh, and by the way, your explanation does not solve this problem, so it is incorrect. And please do not put down another person's efforts. Different people have different levels of comfort with a subject. What takes one person an hour to do, might take someone else a second.
What wonders me is whether the question is concerned about the 'speed' during the journey from earth to space or the final constant velocity? ... Though the later makes more sense since former requires burning of tremendous fuel to maintain the velocity it attains once , and since the object starts from rest on earth , and then gains velocity as it travels higher , maintaining a constant velocity is out of question, as it needs to first attain some velocity before it decides to maintain it. But if the question is concerned about the constant velocity at the end of the process , this is possible , since once in space, it can maintain the velocity it achieved quit easily.
The answer to your question will still remain the (3) option , even if we were not concerned about the CONSTANT VELOCITY thing , because the basic process remains the same , the process is basically an energy-change process.
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