38.9k views
2 votes
If satellite velocity of B is four times the distance of satellite A from the Earth, then the escape satellite B is how many times that of satellite A

a. 1/16
b. 1/4
c. 1/2
d. 2 times
e. 4 times
f. 16 times

1 Answer

3 votes

Final answer:

The question involves concepts of orbital dynamics and escape velocity in physics. However, the problem statement is incomplete or incorrect, as escape velocity is not directly proportional to the distance from Earth, and cannot be determined by satellite velocity alone without other factors such as Earth's mass and gravitational constant.

Step-by-step explanation:

The question refers to orbital dynamics and specifically the concept of escape velocity, which is the speed required for an object to break free from the gravitational attraction of a celestial body without further propulsion.

Escape velocity is not directly proportional to the distance from the Earth, so if satellite B's velocity is four times the velocity of satellite A, we cannot deduce a direct multiple for its escape velocity based on the provided information. Instead, we need to consider the gravitational potential energy and kinetic energy of the satellites. To calculate the escape velocity, a more complex equation that accounts for the mass of the Earth, the gravitational constant, and the distance from the Earth's center (radius of the orbit) is required.

Given this background, the question seems to be incomplete or presents a misunderstanding of the physics involved, as there's no direct proportionality between escape velocity and distance that can be used to solve the problem with the given information.

User Snowball
by
7.6k points