physics homework help needed

Question

The shape of the earth's orbit around the sun is essentially a circle. Assuming the earth maintains a constant speed as it orbits around the sun, which statement about the earth's motion is most correct?

Select one:

a. Its acceleration is zero since it is moving at constant speed.

b. Its acceleration is not zero but is very small.

c. Its acceleration is 9.8 m/s².

d. Its acceleration is 1.0 m/s².

I kinda think that the answer maybe D but i'm not to sure.

Answer 1

Dear experts and students! I thought that it is established fact that the acceleration of the Earth with respect to the Sun has order of magnitude 10 ^{- 3} m/s².

This is easy to prove, if you take regular physics. The problem assumes that the shape of the Earth's orbit is a circle. Hence we can use the formula for centripetal acceleration

a_c = v² / r.

Here v - orbital speed of the Earth, that is easy to find from equation

v = 2πr / T .

r - distance between the Earth and the Sun equal to 1 astronomical unit, or 150 millions km .

This is 1.5 x 10 ¹¹ m. T - the period of 1 revolution of Earth against the Sun equal to 1 year. BTW, for the problems it is nice to remember that in seconds it is approximately T = π x 10⁷ s.

The final formula for centripetal acceleration

a_c = 4π²r / T²

Calculations is easy

a _c = 4 π² 1.5 x 10 ¹¹ / π² x 10¹⁴ = 6 x 10 ^{- 3} m/s²

The right answer is (b).

To the same conclusion we can come using simple logic.

(a) is wrong because any curved motion is accelerated motion.

(c) is not right , because it is acceleration due to gravity for the Earth.

(d) is too big, because although the masses of the Sun and Earth big, the gravitational force between them is inversely proportional to the square of the distance, which is huge - 150 000 000 km.

(b) is left!

Answer 2

This is a clever question. You're right that the answer is not zero. Even if the speed is constant, there is a centripetal or radial acceleration due to the constant change in direction. The equations that relate to centripetal acceleration are a_c=v²/r or a_c=ω²r. (v=2πr/T and ω=2π/T, where T is the orbital period = 365.25 days = 3.16x10⁷s) Now it's just a matter of "Googling" the radius of the Earth's orbit and plugging into either equation (or both to prove to yourself that you did it correctly).

Answer 3

Remember that acceleration is a vector, so as long as the Earth is moving in a circular motion, it is accelerating. We can therefore eliminate answer choice A. I don't see how you could select answer choice C or D without more information, since the problem does not include any numbers. Do they expect you to look up values? If not, then we can safely eliminate those answer choices. (C is a trap answer, since that is the magnitude of acceleration of an object in free fall near the Earth's surface.) B is most likely to be correct, even though "very small" is quite vague.

Answer 4

None of these answers is correct. Even though the speed is constant, the direction is changing (since the earth is not moving in a straight line). For circular motion, the acceleration is the centripetal acceleration,

a_c = v²/r

where v is the orbital speed and r is the radius of the circular orbit. The terms "small" and "big" are very relative (i.e. small or big compared to what?).