Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.
Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
Explanation:
It is given that,
An electron is released from rest in a weak electric field of, 
Vertical distance covered, 
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

.............(1)
Electric force is
and force of gravity is
. As both forces are acting in downward direction. So, total force is:



Acceleration of the electron, 


Put the value of a in equation (1) as :


v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.
The weight changes but the mass will stay the same.
The right answer for the question that is being asked and shown above is that: "C) The alpha particle will be deflected in a curve path. " the result of an alpha particle coming into a magnetic field is that <span>C) The alpha particle will be deflected in a curve path. </span>
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