The earth orbits the sun once every 365.24 days. The distance between the earth and sun is 1.49x 10^8 Km. The mass of the earth is 5.96 x 10 ^24 kg. A) Find the magnitude in newtons of the centripetal force acting on the earth. The mass of mars is 6.56 x 10 ^23 Kg and the distance between Mars and the sun is 2.27 x 10 ^8 km. B) Use Keplers third law to find the period of Mars in Earth days.C) Find the magnitude in Newtons of the centripetal force acting on Mars.
A) To find the magnitude of the centripetal force acting on the Earth, we can use the formula:
F = (m * v^2) / r
Where:
F is the centripetal force,
m is the mass of the object (in this case, the mass of the Earth),
v is the orbital velocity,
and r is the distance between the object and the center of its orbit.
To determine the orbital velocity, we can use the formula:
v = (2 * π * r) / T
Where:
v is the orbital velocity,
π is a constant (approximately 3.14159),
r is the distance between the Earth and the Sun (1.49x10^8 km),
and T is the period of the Earth's orbit (365.24 days, which we need to convert to seconds).
First, let's convert the distance between the Earth and the Sun from kilometers to meters:
r = 1.49x10^8 km * 1000 m/km = 1.49x10^11 m
Next, let's convert the period T from days to seconds:
T = 365.24 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 3.15x10^7 seconds
Now, we can calculate the orbital velocity:
v = (2 * π * 1.49x10^11 m) / (3.15x10^7 seconds)
≈ 2.98x10^4 m/s
Finally, we can substitute the values into the centripetal force formula to find the magnitude of the force:
F = (5.96x10^24 kg * (2.98x10^4 m/s)^2) / 1.49x10^11 m
≈ 3.53x10^22 N
Therefore, the magnitude of the centripetal force acting on the Earth is approximately 3.53x10^22 Newtons.
B) Kepler's third law relates the period of a planet's orbit to its distance from the Sun. The law states:
(T1 / T2)^2 = (r1 / r2)^3
Where:
T1 and T2 are the periods of two planets,
r1 and r2 are the distances of the two planets from the Sun.
To find the period of Mars in Earth days, we can rearrange the equation and solve for T2:
T2 = T1 * (r2 / r1)^(3/2)
We know the period of the Earth (T1) is 365.24 days and the distance between Mars and the Sun (r2) is 2.27x10^8 km. To use this equation, we need to convert the distance to meters and r1 to match the units:
r1 = 1.49x10^11 m (distance between Earth and the Sun)
r2 = 2.27x10^8 km * 1000 m/km = 2.27x10^11 m (distance between Mars and the Sun)
Substituting these values into the equation:
T2 = 365.24 days * (2.27x10^11 m / 1.49x10^11 m)^(3/2)
≈ 686.98 days
Therefore, the period of Mars in Earth days is approximately 686.98 days.
C) To find the magnitude of the centripetal force acting on Mars, we can use the same formula as before:
F = (m * v^2) / r
We will use the mass of Mars (6.56x10^23 kg) and the distance between Mars and the Sun (2.27x10^11 m). However, we need to calculate the orbital velocity (v) for Mars first.
Using the same formula as before:
v = (2 * π * r) / T
Substituting the values for Mars:
v = (2 * π * 2.27x10^11 m) / (686.98 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute)
≈ 2.41x10^4 m/s
Now, we can use the centripetal force formula:
F = (6.56x10^23 kg * (2.41x10^4 m/s)^2) / 2.27x10^11 m
≈ 1.41x10^17 N
Therefore, the magnitude of the centripetal force acting on Mars is approximately 1.41x10^17 Newtons.