QuestionIf Earth spins faster, find the period $T$ for centripetal acceleration $a = 9.8 \, \mathrm{m/s}^{2}$ using $a = \frac{4\pi^{2}r}{T^{2}}$ . Use $r = 6.38 \times 10^{6}$ m. Output $T$ in minutes.

Studdy Solution

STEP 1

Assumptions1. The centripetal acceleration is $9.8 \, \mathrm{m/s}^{}$ . The radius of the Earth is $6.38 \times10^{6}$ m3. The formula for centripetal acceleration is $a = \frac{4\pi^{}r}{^{}}$
4. We need to find the period $$ in minutes

STEP 2

First, we need to rearrange the formula for centripetal acceleration to solve for $$.\[ = \sqrt{\frac{4\pi^{2}r}{a}}\]

STEP 3

Now, plug in the given values for the radius of the Earth and the centripetal acceleration to calculate the period.
$= \sqrt{\frac{\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}}$

STEP 4

Calculate the period $$ in seconds.
$= \sqrt{\frac{4\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}} =5056.\, \mathrm{s}$

STEP 5

Since we need the period in minutes, we need to convert the period from seconds to minutes. We know that1 minute is equal to60 seconds.
$_{\mathrm{min}} = \frac{T_{\mathrm{s}}}{60}$

STEP 6

Plug in the value for the period in seconds to calculate the period in minutes.
$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60}$

STEP 7

Calculate the period in minutes.
$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60} =84.275\, \mathrm{min}$
So, if the Earth were spinning such that the centripetal acceleration is equal to $9. \, \mathrm{m/s}^{2}$ , the period of one full rotation of the Earth would be approximately84.275 minutes.

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QuestionIf Earth spins faster, find the period $T$ for centripetal acceleration $a = 9.8 \, \mathrm{m/s}^{2}$ using $a = \frac{4\pi^{2}r}{T^{2}}$ . Use $r = 6.38 \times 10^{6}$ m. Output $T$ in minutes.

Studdy Solution

STEP 1

Assumptions1. The centripetal acceleration is $9.8 \, \mathrm{m/s}^{}$ . The radius of the Earth is $6.38 \times10^{6}$ m3. The formula for centripetal acceleration is $a = \frac{4\pi^{}r}{^{}}$
4. We need to find the period $$ in minutes

STEP 2

First, we need to rearrange the formula for centripetal acceleration to solve for $$.\[ = \sqrt{\frac{4\pi^{2}r}{a}}\]

STEP 3

Now, plug in the given values for the radius of the Earth and the centripetal acceleration to calculate the period.
$= \sqrt{\frac{\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}}$

STEP 4

Calculate the period $$ in seconds.
$= \sqrt{\frac{4\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}} =5056.\, \mathrm{s}$

STEP 5

Since we need the period in minutes, we need to convert the period from seconds to minutes. We know that1 minute is equal to60 seconds.
$_{\mathrm{min}} = \frac{T_{\mathrm{s}}}{60}$

STEP 6

Plug in the value for the period in seconds to calculate the period in minutes.
$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60}$

STEP 7

Calculate the period in minutes.
$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60} =84.275\, \mathrm{min}$
So, if the Earth were spinning such that the centripetal acceleration is equal to $9. \, \mathrm{m/s}^{2}$ , the period of one full rotation of the Earth would be approximately84.275 minutes.

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QuestionIf Earth spins faster, find the period TTT for centripetal acceleration a=9.8 m/s2a = 9.8 \, \mathrm{m/s}^{2}a=9.8m/s2 using a=4π2rT2a = \frac{4\pi^{2}r}{T^{2}}a=T24π2r​. Use r=6.38×106r = 6.38 \times 10^{6}r=6.38×106 m. Output TTT in minutes.

Studdy Solution

STEP 1

Assumptions1. The centripetal acceleration is 9.8 m/s9.8 \, \mathrm{m/s}^{}9.8m/s . The radius of the Earth is 6.38×1066.38 \times10^{6}6.38×106 m3. The formula for centripetal acceleration is a=4πra = \frac{4\pi^{}r}{^{}}a=4πr​4. We need to find the period $$ in minutes

STEP 2

First, we need to rearrange the formula for centripetal acceleration to solve for $$.\[ = \sqrt{\frac{4\pi^{2}r}{a}}\]

STEP 3

Now, plug in the given values for the radius of the Earth and the centripetal acceleration to calculate the period.=π2×6.38×106 m9.8 m/s2 = \sqrt{\frac{\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}}=9.8m/s2π2×6.38×106m​​

STEP 4

Calculate the period $$ in seconds.=4π2×6.38×106 m9.8 m/s2=5056. s = \sqrt{\frac{4\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}} =5056.\, \mathrm{s}=9.8m/s24π2×6.38×106m​​=5056.s

STEP 5

Since we need the period in minutes, we need to convert the period from seconds to minutes. We know that1 minute is equal to60 seconds.min=Ts60_{\mathrm{min}} = \frac{T_{\mathrm{s}}}{60}min​=60Ts​​

STEP 6

Plug in the value for the period in seconds to calculate the period in minutes.min=5056.5 s60_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60}min​=605056.5s​

STEP 7

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QuestionIf Earth spins faster, find the period $T$ for centripetal acceleration $a = 9.8 \, \mathrm{m/s}^{2}$ using $a = \frac{4\pi^{2}r}{T^{2}}$ . Use $r = 6.38 \times 10^{6}$ m. Output $T$ in minutes.

Assumptions1. The centripetal acceleration is $9.8 \, \mathrm{m/s}^{}$ . The radius of the Earth is $6.38 \times10^{6}$ m3. The formula for centripetal acceleration is $a = \frac{4\pi^{}r}{^{}}$
4. We need to find the period $$ in minutes

Now, plug in the given values for the radius of the Earth and the centripetal acceleration to calculate the period.
$= \sqrt{\frac{\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}}$

Calculate the period $$ in seconds.
$= \sqrt{\frac{4\pi^{2} \times6.38 \times10^{6}\, \mathrm{m}}{9.8\, \mathrm{m/s}^{2}}} =5056.\, \mathrm{s}$

Since we need the period in minutes, we need to convert the period from seconds to minutes. We know that1 minute is equal to60 seconds.
$_{\mathrm{min}} = \frac{T_{\mathrm{s}}}{60}$

Plug in the value for the period in seconds to calculate the period in minutes.
$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60}$