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.