Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v =
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c
Answer:
The ball's initial kinetic energy
The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy
Explanation:
A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.
From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.
Ki+Ui=Kf+Uf
Ki=initial kinetic energy
Ui=initial potential energy
Kf=final kinetic energy
Uf=final potential energy
we know that
m=mass of the ball
ha=downward height a
hb=upward height b
u=initial velocity u
v=final velocity v, which is 0
g=acceleration due to gravity
v=0 at final velocity
1/2mu^2+mgha=0+1/2mv^2
ha=hb+Ki/mh
From the above equation, we can conclude that the ball's initial kinetic energy is responsible for making the ball reach point B.
Point B is higher than point A from the motion gained by the ball