The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 X 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 Mercury (orbit radius = 5.79 X 107 km, orbital period = 88.0 days)
r = 1.5 * 10^11 meters
2 pi r = 9.42 * 10^11 meters circumference
and
365d * 24h/d * 3600 s/h =3.15 * 10^7 seconds
v = 9.42*10^11 m/3.15*10^7 s = 2.99*10^4 m/s
Ac = v^2/r
etc
To find the magnitude of the orbital velocity of the Earth, we can use the equation:
V = (2 * pi * r) / T
where:
V is the orbital velocity
pi is a mathematical constant approximately equal to 3.14
r is the radius of the Earth's orbit around the sun
T is the time taken for one orbit (365 days in this case)
(a) Substituting the values into the equation:
V = (2 * 3.14 * 1.50 X 10^8 km) / (365 days)
Since we need the orbital velocity in m/s, we need to convert kilometers to meters and days to seconds:
1 km = 1000 m
1 day = 24 hours * 60 minutes * 60 seconds = 86,400 seconds
Therefore, we can proceed with the calculation:
V = (2 * 3.14 * 1.50 X 10^8 km * 1000 m/km) / (365 days * 86400 s/day)
Calculating this expression will give us the orbital velocity of the Earth in m/s.
(b) To find the radial acceleration of the Earth toward the sun, we can use the equation:
a = V^2 / r
where:
a is the radial acceleration
V is the orbital velocity
r is the radius of the Earth's orbit around the sun
By substituting the values into the equation, we can calculate the radial acceleration of the Earth toward the sun.
To repeat parts (a) and (b) for the motion of the planet Mercury, we can follow the same procedure using the given values for the orbit radius and orbital period of Mercury and plug them into the equations above.
To find the magnitude of the orbital velocity and the radial acceleration of the Earth and Mercury, we can use the following formulas:
(a) Orbital velocity (v):
- Formula: v = 2πr / T
- Where v is the orbital velocity, r is the radius of the orbit, and T is the orbital period.
(b) Radial acceleration (a):
- Formula: a = v^2 / r
- Where a is the radial acceleration, v is the orbital velocity, and r is the radius of the orbit.
Let's calculate the answers step-by-step:
For the Earth:
(a) Orbital velocity:
- Given: r = 1.50 x 10^8 km = 1.50 x 10^11 m
T = 365 days = 3.15576 x 10^7 seconds (1 year contains 365 days, and each day has 24 hours, each hour has 60 minutes, and each minute has 60 seconds)
- Substituting the values in the formula: v = 2π(1.50 x 10^11) / (3.15576 x 10^7)
- Calculating it: v ≈ 29780 m/s
(b) Radial acceleration:
- Substituting the values in the formula: a = (29780)^2 / (1.50 x 10^11)
- Calculating it: a ≈ 5.88 x 10^-2 m/s^2 (approximately)
For Mercury:
(a) Orbital velocity:
- Given: r = 5.79 x 10^7 km = 5.79 x 10^10 m
T = 88.0 days = 7.6032 x 10^6 seconds
- Substituting the values in the formula: v = 2π(5.79 x 10^10) / (7.6032 x 10^6)
- Calculating it: v ≈ 47990 m/s
(b) Radial acceleration:
- Substituting the values in the formula: a = (47990)^2 / (5.79 x 10^10)
- Calculating it: a ≈ 3.94 x 10^-2 m/s^2 (approximately)
So, the answers are:
For the Earth:
(a) The magnitude of the orbital velocity of the Earth is approximately 29780 m/s.
(b) The radial acceleration of the Earth toward the Sun is approximately 5.88 x 10^-2 m/s^2.
For Mercury:
(a) The magnitude of the orbital velocity of Mercury is approximately 47990 m/s.
(b) The radial acceleration of Mercury toward the Sun is approximately 3.94 x 10^-2 m/s^2.
(a) Well, before I give you the answer, let me ask you this: Did you know that the Earth's orbital velocity is so fast that Usain Bolt would struggle to keep up? It's quite impressive!
Now, to calculate the orbital velocity, we can use the formula:
v = 2πr / T
where v is the orbital velocity, r is the radius, and T is the period.
So, plugging in the values for the Earth:
v = 2π(1.50 X 10^8 km) / (365 days * 24 hours * 60 minutes * 60 seconds)
Calculating this, the magnitude of the Earth's orbital velocity is approximately 29,524 m/s.
(b) Now, let's find the radial acceleration of the Earth toward the sun. This can be found using the formula:
a = v^2 / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the values for the Earth:
a = (29,524 m/s)^2 / (1.50 X 10^8 km)
Calculating this, the radial acceleration of the Earth toward the sun is approximately 0.0059 m/s^2.
(c) Now let's repeat the process for the planet Mercury. Did you know that Mercury is the speed-demon of our solar system? It zooms around the sun faster than a caffeinated cheetah!
For Mercury:
(a) The magnitude of the orbital velocity is found using the same formula as before:
v = 2π(5.79 X 10^7 km) / (88.0 days * 24 hours * 60 minutes * 60 seconds)
Calculating this, the magnitude of Mercury's orbital velocity is approximately 47,878 m/s.
(b) The radial acceleration of Mercury toward the sun can be found using the same formula as before:
a = (47,878 m/s)^2 / (5.79 X 10^7 km)
Calculating this, the radial acceleration of Mercury toward the sun is approximately 3.841 m/s^2.
So, there you have it! The Earth and Mercury both have their own speedy moves around the sun.