Answer:

Explanation:
assume
M= mass of Mars
m=mass of phobos
r=orbital radius
T=period
we can apply F=ma to this orbital motion (considering the cricular motion laws)
where,
and a=rω^2
where ω=
and G is the universal gravitational constant.
G = 6.67 x 10-11 N m2 / kg2

Answer:
<em>The frequency changes by a factor of 0.27.</em>
<em></em>
Explanation:
The frequency of an object with mass m attached to a spring is given as
= 
where
is the frequency
k is the spring constant of the spring
m is the mass of the substance on the spring.
If the mass of the system is increased by 14 means the new frequency becomes
= 
simplifying, we have
= 
= 
if we divide this final frequency by the original frequency, we'll have
==>
÷
==>
x
==> 1/3.742 = <em>0.27</em>
78.4 F because you do 8.00 muliplyed by 9.8