Which of the following has the longest wave length and the lowest frequency
1 answer:
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Answer:
A) F(r) = [12a/(r^(13))] - [6b/(r^(7))]
ii) Graphs are attached
B) Equilibrium Distance = (2a/b)^(1/6)
C) Minimum Energy = b²/4a
D) a = 6.67 x 10^(-138) Jm^(12)
b = 6.41 x 10^(-78) Jm^(6)
Explanation:
I've attached the explanation of A-C alongside the graphs
D) i) From the question, we are to make r equal to the derivative of "r" we got when F(r) = 0 which was r = (2a/b)^(1/6)
Thus, since we are given equilibrium distance as: 1.13 x 10^(-10), hence;
(2a/b)^(1/6) = 1.13 x 10^(-10)m
So, (2a/b)= [1.13 x 10^(-10)]^(6)
a = (b/2)[1.13 x 10^(-10)]^(6)
From earlier, we saw that b²/4a = U(r)
Thus since U(r) = 1.54 x 10^(-18) from the question, b²/4a = 1.54 x 10^(-18)
Putting a = (b/2)[1.13 x 10^(-10)]^(6);
We have;
(b²) / [(4b/2)[1.13 x 10^(-10)]^(6)]] = 1.54 x 10^(-18)
b/2 = [1.13 x 10^(-10)]^(6)] x 1.54 x 10^(-18)
So, b = 6.41 x 10^(-78) Jm^(6)
ii) Putting (6.41 x 10^(-78))² for b in;
a = (b/2)[1.13 x 10^(-10)]^(6)
We have, a = (6.41 x 10^(-78))²/ ( 4 x 1.54 x 10^(-18)
So a = 6.67 x 10^(-138) Jm^(12)
Answer:
The new speed must be
Explanation:
In order for the satellite to be on a stable orbit around the planet, the gravitational attraction must be equal to the centripetal force that keeps the satellite in circular motion:
where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, V0 the speed of the satellite at distance R from the center of the planet.
We can re-write V0, the initial satellite speed, by re-arranging the equation:
Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be: