Question

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).

Answer 1

(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:

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

where

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

$T=365 d \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =3.15 \cdot 10^7 s$ is the time taken for the Earth to complete one orbit

Substituting into the first equation, we find the orbital velocity:

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

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

The radial acceleration of the Earth toward the sun, which corresponds to the centripetal acceleration, is

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

where

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

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

Substituting into the equation, we have

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

(c) 6,801 m/s

For planet Uranus, we have

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

$T=84.02 y \cdot 365 d/y \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =2.65 \cdot 10^9 s$ is the orbital period

So, the orbital velocity is

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

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

For planet Uranus, we have

v = 6,801 m/s is the orbital velocity

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

So, the radial acceleration is

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