Answer:
The acceleration due to gravity on the other planet is 1/4 of the acceleration due to gravity on earth
Explanation:
Using the formula;
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex] ......................equation 1
for earth we have;
[tex]1=2\pi \sqrt{\frac{l}{g_{e} } }[/tex] ..................equation 2
for the other planet we have;
[tex]2=2\pi \sqrt{\frac{l}{g_{p} } }[/tex] ...................equation 3
divide the equation 3 by equation 2;
[tex]\frac{2\pi \sqrt{\frac{l}{g_{p} } }}{2\pi \sqrt{\frac{l}{g_{e} } }} =\frac{2}{1}[/tex]
[tex]\frac{ \sqrt{\frac{l}{g_{p} } }}{\sqrt{\frac{l}{g_{e} } }} =\frac{2}{1}[/tex]
[tex]\sqrt{\frac{l}{g_{p} } }}*{ \sqrt{\frac{g_{e}}{l } }} =\frac{2}{1}[/tex]
[tex]\sqrt{\frac{g_{e} }{g_{p} } } =\frac{2}{1}[/tex]
[tex]\frac{g_{e} }{g_{p} }=\frac{2^{2} }{1^{2} }}[/tex]
[tex]\frac{g_{e} }{g_{p} }=4[/tex]
[tex]g_{p}=\frac{1}{4} g_{e}[/tex]
The gravitational acceleration on the other planet will be one-fourth of the Earth's gravitational acceleration.
The formula for the period of the pendulum,
[tex]T = 2 \pi \sqrt {\dfrac lg}[/tex]
Where,
[tex]T[/tex] - period
[tex]l[/tex]- length
[tex]g[/tex] - gravitational acceleration
On the earth,
[tex]1 = 2 \pi \sqrt {\dfrac l{g_e}}[/tex]......................1
On the other planet,
[tex]2 = 2 \pi \sqrt {\dfrac l{g_o}}[/tex].....................2
From equations 1 and 2,
[tex]\dfrac 21 = \frac { 2 \pi \sqrt {\dfrac l{g_e}}}{2 \pi \sqrt {\dfrac l{g_o}}}\\ 2^2 = \dfrac {g_e}{g_o}\\\\g_o = \dfrac 14 g_e[/tex]
Therefore, the gravitational acceleration on the other planet will be one-fourth of the Earth's gravitational acceleration.
To know more about gravitational acceleration.
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