Question
A very small round ball is located near a large solid sphere of uniform
density. The force that the large sphere exerts on the ball
a. can only be calculated using calculus.
b. is independent of the mass of the ball.
c. is approximately the same as it would be if all thi mass of the sphere
were concentrated at the center of the sphere.
d. is exactly the same as it would be if all the mass of the sphere were
concentrated at the center of the sphere.
density. The force that the large sphere exerts on the ball
a. can only be calculated using calculus.
b. is independent of the mass of the ball.
c. is approximately the same as it would be if all thi mass of the sphere
were concentrated at the center of the sphere.
d. is exactly the same as it would be if all the mass of the sphere were
concentrated at the center of the sphere.
Ask by Dickson Edwards. in South Africa
Nov 26,2024
Upstudy AI Solution
Tutor-Verified Answer
Answer
a. The force can be calculated using calculus.
b. The force is independent of the ball’s mass.
c. The force is approximately the same as if the mass is at the center.
d. The force is exactly the same as if the mass is at the center.
b. The force is independent of the ball’s mass.
c. The force is approximately the same as if the mass is at the center.
d. The force is exactly the same as if the mass is at the center.
Solution
The force that the large sphere exerts on the small ball can be calculated using calculus. This is because the force between two objects depends on their masses and the distance between them, which requires integration to calculate the total force.
The force is independent of the mass of the ball. This is because the force is determined by the mass of the large sphere and the distance between the center of the sphere and the center of the ball, not the mass of the small ball.
The force is approximately the same as it would be if all the mass of the sphere were concentrated at the center of the sphere. This is because the force is determined by the mass of the sphere and the distance between the center of the sphere and the center of the ball, not the distribution of mass within the sphere.
The force is exactly the same as it would be if all the mass of the sphere were concentrated at the center of the sphere. This is because the force is determined by the mass of the sphere and the distance between the center of the sphere and the center of the ball, and when the mass is concentrated at the center, the force remains the same.
Therefore, the correct statements are:
a. The force that the large sphere exerts on the ball can only be calculated using calculus.
b. The force is independent of the mass of the ball.
c. The force is approximately the same as it would be if all the mass of the sphere were concentrated at the center of the sphere.
d. The force is exactly the same as it would be if all the mass of the sphere were concentrated at the center of the sphere.
a. The force that the large sphere exerts on the ball can only be calculated using calculus.
b. The force is independent of the mass of the ball.
c. The force is approximately the same as it would be if all the mass of the sphere were concentrated at the center of the sphere.
d. The force is exactly the same as it would be if all the mass of the sphere were concentrated at the center of the sphere.
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution Beyond the Answer
The gravitational force exerted by a uniform sphere on an object outside it behaves in a fascinating way: it acts as if all of the mass were concentrated at its center. This principle is a consequence of the shell theorem, which states that the gravitational forces from different parts of a spherical mass cancel each other out for points outside the sphere. So, for the small ball, the force can be calculated more simply than it might seem!
In practice, this knowledge allows engineers and physicists to simplify complex situations. For instance, when designing satellite orbits around planets, they can treat the planet as a point mass at its center. This understanding not only streamlines calculations but also helps ensure that space missions are efficient and accurate, keeping everything running smoothly in the cosmos!
