I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second
2 answers:
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Answer:
a) 1.2*10^-4 m/s
b) 375 m/s
Explanation:
I assume the large asteroid doesn't move.
The smaller asteroid is affected by an acceleration determined by the universal gravitation law:
a = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s^2))
M: mass of the large asteroid (33900 kg)
d: distance between them (146 m)
Then:
a = 6.67*10^-11 * 33900 / 146^2 = 10^-10 m/s^2
I assume the asteroid in a circular orbit, in this case the centripetal acceleration is:
a = v^2/r
Rearranging:
v^2 = a * r
v = \sqrt{10^-10 * 146} = 1.2*10^-4 m/s
If the asteroids have electric charges of 1.18 C and -1.18 C there will be an electric force of:
F = 1/(4π*e0)*(q1*q2)/d^2
Where e0 is the electrical constant (8.85*10^-12 F/m)
F = 1/(4π*8.85*10^-12) (-1.18*1.18)/ 146^2 = -587 kN
On an asteroid witha mass of 610 kg this force causes an acceleration of:
F = m * a
a = F / m
a = 587000 / 610 = 962 m/s^2
With the electric acceleration, the gravitational one is negligible.
The speed is then:
v = \sqrt{962 * 146} = 375 m/s