Answer:
3100.05 N
Explanation:
= Force on piston = 45 N
= Radius of piston
= Radius of plunger

From Pascal's law we have

The force is 3100.05 N
Mass of Jupiter=1.9×10
27
㎏=M
1
Mass of Sun=1.99×10
30
㎏=M
2
Mean distance of Jupiter from Sun=7.8×10
11
m=r
G=6.67×10
−11
N㎡㎏
−2
Gravitational Force, F=
r
2
GM
1
M
2
F=
(7.8×10
11
)
2
6.67×10
−11
×1.9×10
27
×1.99×10
30
F=4.16×10
23
N
Answer:
R=4.22*10⁴km
Explanation:
The tangential speed
of the geosynchronous satellite is given by:

Because
is the circumference length (the distance traveled) and T is the period (the interval of time).
Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

If we substitute the expression for
in this formula, we get:

Since the centripetal force is the gravitational force
between the satellite and the Earth, we know that:
![F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }](https://tex.z-dn.net/?f=F_g%3D%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%5C%5C%5C%5C%5Cimplies%20%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%3D%5Cfrac%7B4m%5Cpi%20%5E%7B2%7DR%7D%7BT%5E%7B2%7D%7D%5C%5C%5C%5CR%5E%7B3%7D%3D%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%5C%5C%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%7D)
Where G is the gravitational constant (
) and M is the mass of the Earth (
). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:
![R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7DNm%5E%7B2%7D%2Fkg%5E%7B2%7D%29%285.97%2A10%5E%7B24%7Dkg%29%2886400s%29%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%7D%5C%5C%5C%5CR%3D4.22%2A10%5E%7B7%7Dm%3D4.22%2A10%5E%7B4%7Dkm)
This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.
Answer:
Potential Energy
Explanation:
Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.