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jok3333
20 days ago
9

The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 108 km, and the earth travels around this orbit

in 365 days. (a) What is the magnitude of the orbital velocity of the earth in m/s? (b) What is the radial acceleration of the earth toward the sun in m/s2? (c) Repeat parts (a) and (b) for the motion of the planet Uranus (orbit radius = 2.87 109 km, orbital period = 84.02 years).
Physics
1 answer:
kicyunya [2.2K]20 days ago
6 0

(a) 29,905 m/s

The speed of Earth's orbit around the Sun can be obtained by dividing the total distance of the orbit by the duration of the orbital period:

v=\frac{2\pi r}{T}

where

r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m represents the radius of the orbit

T=365 d \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =3.15 \cdot 10^7 s is the period taken for a complete orbit by Earth

By plugging values into this formula, we can calculate the orbital speed:

v=\frac{2\pi (1.50\cdot 10^{11} m)}{(3.15\cdot 10^7 s)}=29,905 m/s

(b) 5.96\cdot 10^{-3} m/s^2

The centripetal acceleration that Earth experiences as it is drawn toward the sun is called radial acceleration:

a=\frac{v^2}{r}

where

v = 29,905 m/s is the orbital speed

r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m is the radius of the orbit

By inserting into the relevant equation, we can derive:

a=\frac{(29,905 m/s)^2}{(1.50\cdot 10^{11} m)}=5.96\cdot 10^{-3} m/s^2

(c) 6,801 m/s

For the planet Uranus, we can determine:

r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m is the radius of its orbit

T=84.02 y \cdot 365 d/y \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =2.65 \cdot 10^9 s is the time it takes for one complete orbit

Thus, the orbital speed is calculated as:

v=\frac{2\pi (2.87\cdot 10^{12} m)}{(2.65\cdot 10^9 s)}=6,801 m/s

(d) 1.61\cdot 10^{-5} m/s^2

For Uranus, with v = 6,801 m/s marking the orbital speed

r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m is the orbital radius

Thus, the radial acceleration can be expressed as:

a=\frac{(6,801 m/s)^2}{(2.87\cdot 10^{12} m)}=1.61\cdot 10^{-5} m/s^2

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Explanation:

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Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

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r = \frac{R}{n^{1/3}}

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