Physics, 14.01.2020 17:31 paigedice8090

Determine the acceleration of earth due to its motion around the sun. assume the earth's orbit to be circular with a radius of 1.5 x 10^8 km.

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The definition of centripetal acceleration allows to find the result for the acceleration of the earth around the sun is:

a = 5.95 10⁻³ m / s²

Centripetal acceleration is the acceleration of a body where all the energy is used to change the direction of the velocity keeping the value of its modulus constant.

$a = \frac{v^2}{r}$

They indicate that the radius of the circular orbit is r = 1.5 10⁸ km = 1.5 10¹¹m.

In a circular motion the speed is constant, therefore we can use the uniform motion relation to find the speed.

$v= \frac{L}{t}$

The length of the orbital circle is:

L = 2π r

The time for a complete orbit is called the period.

v = $\frac{2\pi r}{T}$

The period of the Earth's orbit is one year.

T = 1 year (365 days / 1 year) (24 h / 1 day) (3600 s = 1 h)

T = 3.154 10⁷ s

Let's calculate the velocity.

v = $\frac{2 \pi \ 1.5 \ 10^{11}}{6.154 \ 10^7}$ 2 pi 1.5 1011 / 3.154 10⁷

v = 2.988 10⁴ m / s

Now we can calculate the acceleration.

$a = \frac{(2.988 \ 10^4)^2 }{1.5 \ 10^{11}}$

a = 5.95 10⁻³ m / s²

In conclusion using the definition of centripetal acceleration we can find the result for the acceleration of the earth around the sun is:

a = 5.95 10⁻³ m / s²

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Answer from: sdolina5

Explanation:

Given

Radius of orbit $r=1.5\times 10^8\ km$

As earth moves around sun it experiences a centripetal force on it which is given by

$a_c=\frac{v^2}{r}$

where v=velocity of earth

Time period of Revolution

$T=\frac{2\pi \cdot r}{v}$

time period is 365 days i.e. $3.153\times 10^7\ s$

$v=\frac{2\pi \times 1.5\times 10^{11}}{3.153\times 10^7}$

$v=2.98\times 10^4\ m/s$

$a_c=\frac{v^2}{r}$

$a_c=\frac{9\times 10^8}{1.5\times 10^{11}}$

$a_c=0.006\ m/s^2$

acceleration due to gravity is $9.8\ m/s^2$

a_c is less than gravity

Determine the acceleration of earth due to its motion around the sun. assume the earth's orbit to be

Answer from: Quest

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Answer from: Quest

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