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Natalka [10]
3 years ago
12

Suppose we imagine the Sun to be about the size of a grapefruit. Which of the following describes the size and distance of Earth

on the same scale? Earth is the size of a ball point about 1 meter away from the Sun. Earth is the size of a golf ball about 1 meter away from the Sun. Earth is the size of a ball point about 15 meters away from the Sun. Earth is the size of a golf ball about 15 meters away from the Sun. Earth is the size of a marble about 25 miles away from the Sun.
Geography
2 answers:
galben [10]3 years ago
4 0

Answer:

'Earth is the size of a tip of a ballpoint pen about 15 m away from the sun, is the correct answer.

Explanation

Radius of sun = 7 x 10^8 m

Radius of the earth = 6.37 x 10^6 m

Radius of grapefruit = 7.6 x 10^-2 m

Since sun is the size of a grapefruit.

So, our scale = (7.6 x 10^-2) / (7 x 10^8) = 10^-10

Means the scale is approximately, 1:10^-10

So, size of earth = 6.37 x 10^6 x 10^-10 = 6.37 x 10^-4 m (Size of ballpoint)

As we know that the distance between the sun and the earth = 14.9 x 10^10 m

So, in our scale, distance between the sun and the earth = 14.9 x 10^10 x 10^-10 = 14.9 m

which is approximately 15 meters.

Whitepunk [10]3 years ago
3 0

Answer:

Earth is the size of a ball point about 15 meters away from the Sun.

Explanation:

Given

D_grapefruit = 0.15m

D_Sun = 1.3927*10⁹m

D_Earth = 12742000 m

D_golf ball = 0.04267 m

Distance between the Sun and the Earth = d = 149.6*10⁹ m

We get the relation between the grapefruit and the Sun as follows

E = D_grapefruit / D_Sun = 0.15m / 1.3927*10⁹m = 1.077*10⁻¹⁰

then

we have

D = d*E =  (149.6*10⁹ m)*(1.077*10⁻¹⁰) ≈ 15 m

On the other hand

D_golf ball < D_ball point

0.04267 m < 0.147 m

then

D_golf ball / D_Earth = 0.04267 m / 12742000 m = 3.35*10⁻⁹

D_ball point / D_Earth = 0.147 m / 12742000 m = 1.15*10⁻⁸

As  3.35*10⁻⁹ < 1.15*10⁻⁸  

we can say that 1.077*10⁻¹⁰ is closer to 1.077*10⁻¹⁰

finally

Earth is the size of a ball point about 15 meters away from the Sun.

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