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3 years ago

How does gravity affect the orbital period of a planet? PLEASEEEE HELPPPP!!!!!

Chemistry

2 answers:

dedylja [7]3 years ago

5 0

The stronger the gravitational force, the shorter the orbital period.

In other words, the greater the gravitational attraction of its sun, the faster the planet must go to stay in the same orbit.

The formula for the orbital period P is

P = 2π√[r^3/(GM)]

where

r = the average radius of the orbit

G = the gravitational constant

M = the mass of the planet’s sun

The important point is that the orbital period is inversely proportional to the square root of the sun’s mass.

The gravitational attraction of the sun is directly proportional to its mass.

Thus, as mass increases, gravitational attraction increases, and the orbital period decreases

Nat2105 [25]3 years ago

3 0

Gravity holds the earth together with gravity because of the earth's axis the earth spins causing seasons. And that also effects the period of orbit.

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

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Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

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From first source,

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$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

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$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

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Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

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Thus, we can conclude that work function of the metal is 2 eV.

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