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borishaifa [10]
2 years ago
11

Saturn’s moon titan has a mass of 1.35 × 10^23 kg. if titan is 1.19 × 10^6 km from saturn, and saturn’s mass is 5.86 × 1026 kg,

what is the gravitational force between saturn and its moon?
Physics
1 answer:
prohojiy [21]2 years ago
6 0

Answer:The gravitational force equation is Fg=(G*M*m)/r² where, G is the gravitational constant, G=6.67*10^-11 m³/kg*s², M is the mass of Saturn, M=5.86*10^26, m is the mass of Titan, m=1.35*10^23 and r is the distance, r=1.19*10^6 km=1.19*10^9 m. Now we simply input the numbers into the equation:

Fg=(6.67*10^-11)*(5.86*10^26)*(1.35*10^23)/(1.19*10^6)²

Fg=(5.277*10^39)/(1.41*10^18)=3.743*10^21 N  

The correct answer is the third one.

Explanation:

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Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

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