91影视

• The mass of the object is m.
• The distance of the object from the earth is r.
• The magnitude of the shaft is$F_{g}G{m}_E{m}_r/R_E^3$.

The law states that the mass鈥檚 force is directly proportional to the masses鈥� products and also it is inversely proportional to the masses鈥� square. Moreover, the velocity of a body can be identified by the square root of the acceleration and radius in the simple harmonic motion of the body.

The difference in the potential energy can be identified with the help of the product of the masses and divided by the square of the distances amongst them. The speed of the object can be deduced by multiplying the acceleration and radius of the earth and getting the square root of it.

a) From the given formula, the expression for the gravitational potential energy can be expressed as:

$F_g=G\frac{m_E{m}_R}{R_E^3}$

Here, G is the gravitational constant,$m_E$and$m_R$are the mass of the shaft and the object respectively, and$R_E$is the radius of the shaft.

The free-body diagram of the shaft has been illustrated below-

Here, r is the radius of the shaft, the density inside the shaft is constant and the mass of the earth is described as$M_E$.

Hence, as the direction of the object is opposite to that of the shaft, the force expressed by the shaft is expressed as:

$F_g=G\frac{m_E{m}_R}{R_E^3}$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�..(1)

Using the equation (1) the expression for the gravitational potential energy can be expressed as,

$$\begin{gathered} U\left(r\right)-U\left(0\right)=-{鈭�}_r^{r\text{'}}{F}_{g}dr \\ =G\frac{m_E{m}_R}{2{\left(R_E\right)}^3}{r}^2 \end{gathered}$$

Thus, the expression for the gravitational potential energy is$G\frac{m_E{m}_R}{2{\left(R_E\right)}^3}$.

b) From the simple harmonic motion, the formula of the speed of the body can be expressed as:

$$\begin{gathered} v=rw \\ =R\sqrt{\frac{g}{R}} \\ =\sqrt{gR} \\ =\sqrt{9.8m/s^2脳6371脳{10}^{3}m} \\ =7901.6m/s \end{gathered}$$

Thus, the speed of the object when it reaches the center of the earth is$7901.6m/s$