The Sun exerts a gravitational force of 3.56 x 10^22 N on Earth. The mass of the Sun is 2.0 x 10^30 kg. The mass of the Earth is 6.0 x 10^24 kg. If we assume the Earth has a circular orbit of 1.5 x 10^11 m, what is the velocity needed for Earth to remain in orbit?
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To determine the velocity needed for Earth to remain in orbit around the Sun, we can use the concept of centripetal force. The centripetal force acting on an object moving in a circular path is equal to the gravitational force between the object and the other massive body it is orbiting.
Here's how we can calculate it step by step:
Step 1: Calculate the gravitational force between the Sun and Earth using Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force,
G is the gravitational constant (6.67430 x 10^-11 Nm^2/kg^2),
m1 is the mass of the Sun,
m2 is the mass of the Earth,
and r is the distance between the centers of the two bodies.
Let's calculate it:
F = (6.67430 x 10^-11 Nm^2/kg^2) * [(2.0 x 10^30 kg) * (6.0 x 10^24 kg)] / (1.5 x 10^11 m)^2
Step 2: Now, we know that the centripetal force acting on Earth is equal to the gravitational force between the Sun and Earth:
F = m * v^2 / r
Where:
F is the gravitational force,
m is the mass of the Earth,
v is the velocity of the Earth,
and r is the distance between the centers of the two bodies.
Since these forces are equal, we can set them equal to each other and solve for v:
F = m * v^2 / r
F = (6.67430 x 10^-11 Nm^2/kg^2) * [(2.0 x 10^30 kg) * (6.0 x 10^24 kg)] / (1.5 x 10^11 m)^2 = 3.56 x 10^22 N
So now we have:
3.56 x 10^22 N = (6.0 x 10^24 kg) * v^2 / (1.5 x 10^11 m)
Rearranging the equation to solve for v:
v^2 = (3.56 x 10^22 N) * (1.5 x 10^11 m) / (6.0 x 10^24 kg)
v^2 = 8.9 x 10^5 m^2/s^2
Taking the square root of both sides to solve for v:
v = √(8.9 x 10^5 m^2/s^2)
Now, calculating the value of v:
v ≈ 2.98 x 10^4 m/s
Therefore, the velocity needed for Earth to remain in orbit around the Sun is approximately 2.98 x 10^4 m/s.