16. At a particular point, a satellite in an elliptical orbit has a gravitational potential energy of 7000 MJ with respect to Ea
1 answer:
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Answer: V = 504m/a
F = 4N
Explanation: please find the attached file for the solution
Answer:
Explanation:
(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.
(b) Yes, it does and the formula for force of gravity between any two object is
F = G
where m1 and m2 are masses of the first and second object respectively
r is the distance between the center of the two masses
G is the gravitational constant
(a)
The total mechanical energy of the space probe must be constant, so we can write:
(1)
where
is the kinetic energy at the surface, when the probe is launched
is the gravitational potential energy at the surface
is the final kinetic energy of the probe
is the final gravitational potential energy
Here we have
at the surface, (radius of the planet),
(mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
At the final point, the distance of the probe from the centre of Zero is
so the final potential energy is
So now we can use eq.(1) to find the final kinetic energy:
(b)
The probe reaches a maximum distance of
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
At that point, the gravitational potential energy is
So now we can use eq.(1) to find the initial kinetic energy: