The relationship between the two times is [tex]TE = TM\sqrt{\frac{gM}{gE} }[/tex]
The given parameters:
the acceleration due to gravity on the Moon = gM
height from which the object was dropped on moon, = H
time taken for the object to reach the moon surface, = TM
height from which the object was dropped on Earth, = H
the acceleration due to gravity on the Earth = gE
time taken for the object to reach the Earth's surface = TE
To find:
- The relationship between the two times:
Apply the following kinematic equation:
[tex]H = ut + \frac{1}{2} gt^2\\\\we \ know \ nothing \ about \ the \ initial \ velocity \ u , let \ u = 0\\\\H = \frac{1}{2} gt^2\\\\[/tex]
Height equation for the object on the moon:
[tex]H = \frac{(gM) TM^2}{2}[/tex]
Height equation for the object on the Earth:
[tex]H = \frac{(gE) TE^2}{2}[/tex]
Since the two heights are equal, set the two equation to be equal to each other to determine the relationship between the two times:
[tex]\frac{(gE)TE^2}{2} = \frac{(gM)TM^2}{2} \\\\(gE)TE^2 = (gM)TM^2\\\\make \ TE \ the \ subject \ of \ the \ formula;\\\\ TE^2 = \frac{(gM)TM^2}{gE} \\\\TE= \sqrt{\frac{(gM)TM^2}{gE} } \\\\TE = TM\sqrt{\frac{gM}{gE} }[/tex]
Thus, the relationship between the two times is [tex]TE = TM\sqrt{\frac{gM}{gE} }[/tex]
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