Respuesta :
(a) 29,905 m/s
The magnitude of the Earth's orbital velocity around the Sun is given by the circumference of the orbit divided by the time taken:
[tex]v=\frac{2\pi r}{T}[/tex]
where
[tex]r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m[/tex] is the orbital radius
[tex]T=365 d \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =3.15 \cdot 10^7 s[/tex] is the time taken for the Earth to complete one orbit
Substituting into the first equation, we find the orbital velocity:
[tex]v=\frac{2\pi (1.50\cdot 10^{11} m)}{(3.15\cdot 10^7 s)}=29,905 m/s [/tex]
(b) [tex]5.96\cdot 10^{-3} m/s^2[/tex]
The radial acceleration of the Earth toward the sun, which corresponds to the centripetal acceleration, is
[tex]a=\frac{v^2}{r}[/tex]
where
v = 29,905 m/s is the orbital velocity
[tex]r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m[/tex] is the orbital radius
Substituting into the equation, we have
[tex]a=\frac{(29,905 m/s)^2}{(1.50\cdot 10^{11} m)}=5.96\cdot 10^{-3} m/s^2[/tex]
(c) 6,801 m/s
For planet Uranus, we have
[tex]r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m[/tex] is the orbital radius
[tex]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[/tex] is the orbital period
So, the orbital velocity is
[tex]v=\frac{2\pi (2.87\cdot 10^{12} m)}{(2.65\cdot 10^9 s)}=6,801 m/s [/tex]
(d) [tex]1.61\cdot 10^{-5} m/s^2[/tex]
For planet Uranus, we have
v = 6,801 m/s is the orbital velocity
[tex]r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m[/tex] is the orbital radius
So, the radial acceleration is
[tex]a=\frac{(6,801 m/s)^2}{(2.87\cdot 10^{12} m)}=1.61\cdot 10^{-5} m/s^2[/tex]
The acceleration of the earth is [tex]1.992\times 10^{11} \rm\ m/s^2[/tex] while the acceleration of the Uranus is [tex]1.6138\times 10^{-5} \rm\ m/s^2[/tex].
Given to us
The radius of the earth's orbit around the sun (assumed to be circular), r = 1.50 x 10⸠km = [tex]1.50\times 10^{11} \rm\ m[/tex]
Number of days taken by earth to cover the distance, t = 365 days = [tex]365\times 24\times 60\times 60 = 31.536\times10^6[/tex]
What is the magnitude of the orbital velocity of the earth in m/s?
We know the magnitude of the earth's orbital velocity around the sun is given as the circumference of the orbit divided by the time taken,
[tex]\rm Velocity= \dfrac{2\pi r}{time}[/tex]
Substitute the values,
[tex]\rm Velocity= \dfrac{2 \times \pi \times 1.50 \times10^{11}}{31.536\times 10^6}[/tex]
velocity of the earth, v = 29885.77486 m/s
Thus, the velocity of the earth is 29885.77486 m/s.
What is the radial acceleration of the earth toward the sun in m/s²?
The radial acceleration of the earth towards the sun, corresponding to centripetal acceleration,
[tex]a = \dfrac{v^2}{r}[/tex]
[tex]a = \dfrac{29885.77486^2}{1.50\times 10^{11}}\\\\a = 1.992\times 10^{11} \rm\ m/s^2[/tex]
Thus, the acceleration of the earth is [tex]1.992\times 10^{11} \rm\ m/s^2[/tex].
What is the magnitude of the orbital velocity of the Uranus in m/s?
We know the magnitude of Uranus' orbital velocity around the sun is given as the circumference of the orbit divided by the time taken,
[tex]\rm Velocity= \dfrac{2\pi r}{time}[/tex]
Substitute the values,
[tex]\rm Velocity= \dfrac{2 \times \pi \times 2.87 \times10^{12}}{84.02\times 365\times24\times3600}[/tex]
velocity of the Uranus, v = 6805.695 m/s
Thus, the Uranus of the earth is 6805.695 m/s.
What is the radial acceleration of the Uranus toward the sun in m/s²?
The radial acceleration of the Uranus towards the sun, corresponding to centripetal acceleration,
[tex]a = \dfrac{v^2}{r}[/tex]
[tex]a = \dfrac{6805.695 ^2}{2.87\times 10^{12}}\\\\a = 1.6138\times 10^{-5} \rm\ m/s^2[/tex]
Thus, the acceleration of the Uranus is [tex]1.6138\times 10^{-5} \rm\ m/s^2[/tex].
Learn more about Centripetal acceleration:
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