Final answer:
The centripetal acceleration of Earth in its orbit is 5.93 x 10^-3 m/s^2, and the centripetal force acting on Earth is approximately 3.55 x 10^22 N. This force is provided by the gravitational attraction between the Earth and the Sun.
Step-by-step explanation:
To calculate the centripetal acceleration of the Earth as it orbits the Sun, we use the formula for centripetal acceleration ac = v2/r, where v is the orbital velocity and r is the radius of the orbit. The average orbital velocity v can be found using the circumference of the Earth's orbit (2πr) divided by the orbital period (one year, approximated as 3.16 × 107 seconds).
This yields v = 2π(1.5 x 1011 m) / (3.16 × 107 s)
≈ 2.98 x 104 m/s.
Plugging the values into the centripetal acceleration formula gives ac = (2.98 x 104 m/s)2 / (1.5 x 1011 m)
= 5.93 x 10-3 m/s2.
To find the centripetal force acting on the Earth, we use Newton's second law, F = ma, where m is the mass of the Earth and a is the acceleration. Using Earth's mass of 5.98 x 1024 kg and the centripetal acceleration we just calculated, we get F = 5.98 x 1024 kg × 5.93 x 10-3 m/s2
≈ 3.55 x 1022 N.
The force providing this centripetal force is the gravitational force between the Sun and Earth. According to Newton's law of universal gravitation, any two bodies attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In Earth's orbit, this gravitational force acts as the centripetal force that keeps Earth in a nearly circular orbit around the Sun.