# Students used pieces of string to model the orbits of the inner planets. They measured the lengths of

Students used pieces of string to model the orbits of the inner planets. They measured the lengths of the strings and recorded the values in a table. Which table correctly shows the relationship between the positions of the planets and the relative lengths of their orbits? A)
Mercury : 25 CM
Venus: 25 CM
Earth: 25 CM
Mars: 25 CM
B)
Mercury : 25 CM
Venus: 22 CM
Earth: 18 CM
Mars: 15 CM
C)
Mercury : 15 CM
Venus: 22 CM
Earth: 18 CM
Mars: 25 CM
D)
Mercury : 15 CM
Venus: 18 CM
Earth: 22 CM
Mars: 25 CM

## This Post Has 8 Comments

1. maddy3lizabeth says:

Mercury and Venus.

The order of the planets is

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, (and I guess based on your beliefs) and Pluto.

2. needhelpasap8957 says:

1. The distance between the perihelion and the aphelion is 116 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Step-by-step explanation:

Let us revise the equation of the ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where the major axis is parallel to the x-axis

The length of the major axis is 2aThe coordinates of the vertices are (± a , 0)The coordinates of the foci are (± c , 0) , where c² = a² - b²

∵ The Sun is located at a focus of the ellipse

∴ The sun located ate c

∵ The perihelion is the point in a planet’s orbit that is closest to the

Sun ( it is the endpoint of the major axis that is closest to the Sun )

∴ The perihelion is located at the vertex (a , 0)

∵ The closest Mercury comes to the Sun is about 46 million miles

∴ The distance between a and c is 46 million miles

∵ The aphelion is the point in the planet’s orbit that is furthest from

the Sun ( it is the endpoint of the major axis that is furthest from

the Sun )

∴ The aphelion is located at the vertex (-a , 0)

∵ The farthest Mercury travels from the Sun is about 70 million miles

∴ The distance from -a to c is 70 million miles

∴ The distance between the perihelion and the aphelion =

70 + 46 = 116 million miles

1. The distance between the perihelion and the aphelion is 116 million miles

∵ The distance between the perihelion and the aphelion is the

length of the major axis of the ellipse

∵ The length of the major axis is 2 a

∴ 2 a = 116

- Divide both sides by 2

∴ a = 58

∴ The distance from the center of Mercury’s elliptical orbit to the

closest end point to the sun is 58 million miles

∵ The distance between the sun and the closest endpoint is

46 million miles

∴ The distance from the center of Mercury’s elliptical orbit and

the Sun = 58 - 46 = 12 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

∵ The major axis runs horizontally

∴ The equation is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

∵ a = 58

∵ c is the distance from the center to the focus of the ellipse

∴ c = 12

∵ c² = a² - b²

∴ (12)² = (58)² - b²

- Add b² to both sides

∴ (12)² + b² = (58)²

- Subtract (12)² from both sides

∴ b² = (58)² - (12)² = 3220

- Substitute these values in the equation

∴ $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

The eccentricity (e) of an ellipse is the ratio of the distance from the

center to the foci (c) and the distance from the center to the

vertices (a) ⇒ $e=\frac{c}{a}$

∵ c = 12

∵ a = 58

∴ $e=\frac{12}{58}$ = 0.207

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

If the eccentricity is zero, it is not squashed at all and so remains a circle.

If it is 1, it is completely squashed and looks like a line

∵ The eccentricity of the ellipse is 0.207

∵ This number is closed to zero than 1

∴ The shape of the ellipse is near to the shape of the circle

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

#LearnwithBrainly

3. 4804053670 says:

b

Explanation:

4. siriuskitwilson9408 says:

Mercury

1) What is the orbit of the Mercury?

The shape of the orbit is oval

2) Is the Sun at the center of the Earth’s orbit?

No

3) Describe the motion of the Earth throughout its orbit?

Starts fast then gets slower then begins to speed up again

4) Does it move at constant speed?

No

Earth

1) What is the orbit of the Earth?

The shape of the orbit is circular

2) Is the Sun at the center of the Earth’s orbit?

Yes

3) Describe the motion of the Earth throughout its orbit?

Travels at a steady consistent speed.

4) Does it move at constant speed?

Yes

Mars

1) What is the orbit of the Mars?

The shape of the orbit is circular

2) Is the Sun at the center of the Earth’s orbit?

Yes

3) Describe the motion of the Earth throughout its orbit?

Travels at a steady consistent speed.

​4) Does it move at constant speed?

Yes

Saturn

1) What is the orbit of the Saturn?

The shape of the orbit is circular

2) Is the Sun at the center of the Earth’s orbit?

Yes

3) Describe the motion of the Earth throughout its orbit?

Travels at a steady consistent speed.

​4) Does it move at constant speed?

Yes

Neptune

1) What is the orbit of the Neptune?

The shape of the orbit is circular

2) Is the Sun at the center of the Earth’s orbit?

Yes

3) Describe the motion of the Earth throughout its orbit?

Travels at a steady consistent speed.

4) Does it move at constant speed?

Yes

Comet

1) What is the orbit of the Comet?

The shape of the orbit is oval

2) Is the Sun at the center of the Earth’s orbit?

No

3) Describe the motion of the Earth throughout its orbit?

starts slow and gets faster and faster

4) Does it move at constant speed?

No

5. saucybby says:

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Explanation:

6. DepressionCentral says:

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You put your one foot in
You put your one foot out
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7. luna563 says:

The answer is B.mars and venus.

8. gabypinskyb7364 says:

(1) 83.764 million miles

(2) 52.766 million miles

(3) $\frac{x^2}{4900}+\frac{y^2}{2116}=1.$

Step-by-step explanation:

Let the origin C(0,0) be the center of the elliptical path as shown in the figure, where the location of the sun is at one of the two foci, say f.

The standard equation of the ellipse having the center at the origin is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\;\cdots(i)$

where $a$ and $b$ are the semi-axes of the ellipse along the x-axis and y-axis respectively.

Let the points P and A represent the points of perihelion (nearest to the sun) and the aphelion (farthest to the sun) of the closest planet Mercury.

Given that,

CP=46 million miles and

CA=70 million miles.

So, $CP=b$ is the semi-minor axis and $CA=a$ is the semi-major axis.

Let the distances on the axes are in millions of miles. So, the coordinates of the point P and A are $P(0,46)$ and $A(70,0)$ respectively.

(1) From the distance formula, the distance between the perihelion and the aphelion is

$PA=\sqrt{(0-70)^2+(46-0)^2}=83.764$ million miles.

(2) Location of the Sun is at focus, $f$, of the elliptical path.

From the standard relation, the distance of the focus from the center of the ellipse, c, is

$c=ae\;\cdots(ii)$

where $a$ and $e$ are the semi-major axis and the eccentricity of the ellipse.

The eccentricity of the ellipse is

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{46^2}{70^2}}=0.7538$.

Hence, from the equation (i) the distance of the Sun from the center of the elliptical path of the Mercury is

$c=70\times0.7538=52.766$ million miles.

(3) From the equation (i), the equation of the elliptical orbit of Mercury is

$\frac{x^2}{70^2}+\frac{46^2}{b^2}=1$

$\Rightarrow \frac{x^2}{4900}+\frac{y^2}{2116}=1.$

$The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical$