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4 years ago

How are the outer planets different from the inner planets of our solar system?

Physics

2 answers:

andrew11 [14]4 years ago

7 0

Answer:

There are eight planets in the solar system. Four of them are inner and other four are outer planets.

→Outer planets lie beyond asteroid belt.

→Outer planets are gaseous planets and are referred as Jovian planets. Inner planets are terrestrial planets.

→Outer planets have many number of moons where as inner planets have zero to two moons.

→Inner planets lie closer to the Sun as compared to Outer planets.

→Outer planets are larger in size than inner planets.

→ Outer planets have larger orbital period than inner planets.

→Outer planets have ring system unlike inner planets.

Verizon [17]4 years ago

5 0

Answer:

The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars. After an asteroid belt comes the outer planets, Jupiter, Saturn, Uranus and Neptune.

Explanation:

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Answer:

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Explanation:

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3 years ago

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3 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$