Centripetal acceleration of Earth around Sun

So well done!In summary, the Earth's orbit around the sun has a radius of 1.5e11 m and takes one year to complete. Using the equations for centripetal acceleration and period, the centripetal acceleration of the Earth is calculated to be 0.005954 m/s^2. An alternative method using the sun's gravitational force also yields a similar answer of 5.8987E-03 m/s^2. These calculations confirm that the initial answer was correct.
  • #1
blue5t1053
23
1
Problem:
The Earth's orbit (assumed circular) around the sun is 1.5e11 m in radius, and it makes this orbit once a year. What is the centripetal acceleration of the earth?

Equations:
a = (v^2) / r
T = (2*pi*r) / v

My work:
T = (2*pi*r) / v;

1 year = (365 days / 1 year)*(24 hours / 1 day)*(60 mins / 1 day)*(60 secs / 1 min) = 3.1536e7 s;

3.1536e7 s = (2*pi*1.5e11 m) / v;
algebraically rearranged is: v = (2*pi*1.5e11 m) / (3.1536e7 s)
v = 29885.8 m/s

a = (v^2) / r;

a = ((29885.8 m/s)^2) / (1.5e11 m);
a = 0.005954 m/s^2 MY ANSWER

My question is if this is correct? I've been bombarded with tough questions up until this one and I am curious to know if I solved this correctly. It 'seemed' too easy. Confirmation on the answer would be appreciated since I can't find any information on presumed circular rotation around the sun. Thank you.
 
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  • #2
Looks OK to me.
 
  • #3
I haven't plugged the numbers in but your working is correct.
 
  • #4
The answer looks right. An alternative way of solving this question (to check your answer) would be to just ask what is the centripetal acceleration of Earth around the sun, given the sun's gravitation force at our distance.

acceleration=G*m1/r(2)

G=gravitational constant=6.67E-11 m(3)kg(-1)s(-2)
m1=mass of sun (1.00 E30) kg
r=distance to the sun = 1.5E11 m

Ie. acceleration = 6.67E-11 m(3)kg(-1)s(-2) * (1.00 E30) m / [ (1.5E11 m) * (1.5E11 m)]

Answer = 5.8987E-03 ms(-2)

My mass of distance were approximations, but the answer is very close indeed.
 
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FAQ: Centripetal acceleration of Earth around Sun

What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is calculated using the equation a = v^2/r, where v is the velocity of the object and r is the radius of the circle.

How does centripetal acceleration affect the Earth's orbit around the Sun?

The centripetal acceleration of the Earth around the Sun is what keeps the Earth in its orbit. The gravitational force of the Sun acts as the centripetal force, pulling the Earth towards the center of the orbit. This balance between centripetal acceleration and gravitational force allows the Earth to maintain a stable orbit around the Sun.

What is the value of centripetal acceleration for the Earth's orbit around the Sun?

The value of centripetal acceleration for the Earth's orbit around the Sun is approximately 0.0059 m/s^2. This value can be calculated using the equation a = v^2/r, where v is the Earth's orbital velocity (29.78 km/s) and r is the distance between the Earth and the Sun (149.6 million km).

Does the centripetal acceleration of the Earth around the Sun change?

No, the centripetal acceleration of the Earth around the Sun remains constant as long as the Earth's orbital velocity and distance to the Sun remain constant. However, the gravitational force of the Sun can vary slightly due to factors such as the positions of other planets, which can affect the Earth's orbit and therefore its centripetal acceleration.

What other factors can affect the centripetal acceleration of the Earth's orbit around the Sun?

The centripetal acceleration of the Earth's orbit around the Sun can also be affected by factors such as the mass and distance of other planets in the solar system, as well as the shape and orientation of the Earth's orbit. These factors can cause small variations in the Earth's orbital velocity and distance from the Sun, ultimately affecting the centripetal acceleration of the Earth's orbit.

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