The acceleration due to gravity is g/4
The acceleration above the earth surface is given by the relation
g^'=gr^2/〖(h+r)〗^2
Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore
g^'=(gr^2)/〖(r+r)〗^2 =g/4
Thus the acceleration due to gravity on the satellite is g/4.
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;

k = 1.4

Work done is given as;

inlet velocity is negligible;

Therefore, the exit velocity is 629.41 m/s
Answer:
Explanation:
Use the following equation:
and solve for F:
and filling in:

F = 4.0 ×
N