Assuming the Earth revolves around the Sun once every 365.3 days in a
perfectly circular orbit, calculate:
the radius of the Earths orbit if it revolves around the Sun at the
speed of 3.0 × 104 m/s
jhm bh
as always, distance (circumference) = speed * time
then divide that by 2π
To calculate the radius of the Earth's orbit if it revolves around the Sun at a speed of 3.0 × 10^4 m/s, we can make use of the formula for the centripetal force:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the object (in this case, the Earth)
v is the velocity of the object (in this case, 3.0 × 10^4 m/s)
r is the radius of the orbit
The centripetal force in this case is the gravitational force between the Sun and the Earth:
F = G * (m * M) / r^2
Where:
G is the gravitational constant
M is the mass of the Sun
Setting these two equations equal to each other, we have:
G * (m * M) / r^2 = (m * v^2) / r
Since we are trying to solve for r, we can rearrange the equation as follows:
G * M = v^2 * r
Now, let's plug in the given values:
G = 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2) (gravitational constant)
M = 1.989 × 10^30 kg (mass of the Sun)
v = 3.0 × 10^4 m/s
Plugging in these values, we get:
(6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) * (1.989 × 10^30 kg) = (3.0 × 10^4 m/s)^2 * r
Now, we can solve for r by rearranging the equation:
r = [(6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) * (1.989 × 10^30 kg)] / [(3.0 × 10^4 m/s)^2]
Evaluating this equation will give you the radius of the Earth's orbit around the Sun.