In deep space, sphere A of mass MA is located at the origin of an x axis and sphere B of mass MB is located on the axis at a poi
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With the given values of , we have
Try dealing with the powers of 10 first: On the right, we have
Meanwhile, the other values on the right reduce to
Then taking units into account, we end up with the equation
Now we solve for :
or, if taking significant digits into account,
To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.
In the given problem half of the period is equivalent to 1 second so the pendulum period is,
From the equations describing the period of a simple pendulum you have to
Where
g= gravity
L = Length
T = Period
Re-arrange to find L we have
Replacing the values,
In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,
In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.