Given:
density of air at inlet,
density of air at inlet,
Solution:
Now,
(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:
= 1.14
% increase in velocity = =114%
which is 14% more
Therefore % increase in velocity is 14%
Answer:
v = √2G / R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1
/ R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0