1) ![6.67\cdot 10^{-4} N](https://tex.z-dn.net/?f=6.67%5Ccdot%2010%5E%7B-4%7D%20N)
The force of gravitation between the two objects is given by:
![F=G\frac{m_1 m_2}{r^2}](https://tex.z-dn.net/?f=F%3DG%5Cfrac%7Bm_1%20m_2%7D%7Br%5E2%7D)
where
is the gravitational constant
m1 = 20,000 kg is the mass of the first object
m2 = 12,500 kg is the mass of the second object
r = 5 m is the distance between the two objects
Substituting the numbers inside the equation, we find
![F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N](https://tex.z-dn.net/?f=F%3D%286.67%5Ccdot%2010%5E%7B-11%7D%29%5Cfrac%7B%2820%2C000%20kg%29%2812%2C500%20kg%29%7D%7B%285%20m%29%5E2%7D%3D6.67%5Ccdot%2010%5E%7B-4%7D%20N)
2) ![2.7\cdot 10^{-3} N](https://tex.z-dn.net/?f=2.7%5Ccdot%2010%5E%7B-3%7D%20N)
From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:
![F \sim \frac{1}{r^2}](https://tex.z-dn.net/?f=F%20%5Csim%20%5Cfrac%7B1%7D%7Br%5E2%7D)
this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor
![F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F](https://tex.z-dn.net/?f=F%27%5Csim%20%5Cfrac%7B1%7D%7B%28r%2F2%29%5E2%7D%3D4%20%5Cfrac%7B1%7D%7Br%5E2%7D%3D4F)
So, the gravitational force increases by a factor 4. Therefore, the new force will be
![F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N](https://tex.z-dn.net/?f=F%27%20%3D%204%20F%3D4%286.67%5Ccdot%2010%5E%7B-4%7D%20N%29%3D2.7%5Ccdot%2010%5E%7B-3%7D%20N)
3) 12.5 Nm
The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:
![\tau=Fd](https://tex.z-dn.net/?f=%5Ctau%3DFd)
Where, in this case:
F = 25 N is the perpendicular force
d = 0.5 m is the distance between the force and the center
By using the equation, we find
![\tau=(25 N)(0.5 m)=12.5 Nm](https://tex.z-dn.net/?f=%5Ctau%3D%2825%20N%29%280.5%20m%29%3D12.5%20Nm)
4) 0.049 kg m^2/s
The relationship between angular momentum (L), moment of inertia (I) and angular velocity (
) is:
![L=I\omega](https://tex.z-dn.net/?f=L%3DI%5Comega)
In this problem, we have
![I=0.007875 kgm^2](https://tex.z-dn.net/?f=I%3D0.007875%20kgm%5E2)
![\omega=6.28 rad/s](https://tex.z-dn.net/?f=%5Comega%3D6.28%20rad%2Fs)
So, the angular momentum is
![L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s](https://tex.z-dn.net/?f=L%3DI%5Comega%3D%280.007875%20kgm%5E2%29%286.28%20rad%2Fs%29%3D0.049%20kg%20m%5E2%2Fs)