## Related questions with answers

A kilogram of water on the side of Earth nearest the moon is gravitationally attracted to the moon with a greater force is gravitationally attracted to the moon with a greater force than a kilogram of water on the side of Earth farthest from the moon. The difference in force per mass, the tidal force, is approximated by

$F _ { T } = \frac { 4 G M R } { d ^ { 3 } }$

where G is the gravitational constant, M is the mass of the moon, R is the radius of Earth, and d is the distance between the centers of the moon and Earth. Even a 1-kg melon held above your head exerts a tidal force on you. The above equation, however, is a good approximation only when d is much larger than R. The distance d between the center of the melon and your center is only about twice R. So the general tidal force equation (the source of the above equation) must be used, which is

$F _ { \mathrm { T } } = ( 4 G M d R ) / \left( d ^ { 2 } - R ^ { 2 } \right) ^ { 2 }$

Use this to calculate the melon‘s

$F_T$

on you. Here M = 1 kg, and if you’re 2 m tall, R = 1 m, and d = 2 m. How does this compare with the tidal force of the moon on you?

Solution

Verified$\textbf{Given:}$

$F_{T,M} = 3.4 \times 10^{-13} \dfrac{\text{N}}{\text{kg}}$ (Moon's tidal force on you)

$M = 1$ kg

$R = 1$ m

$d = 2$ m

The tidal force can be described by the following formula:

$\begin{gather} F_T = \dfrac{4GMdR}{(d^2 - R^2)^2} \end{gather}$

Using Eq (1), we calculate for the tidal force of the Earth on you as follows:

$\begin{align*} F_T &= \dfrac{4GMdR}{(d^2 - R^2)^2} \\ &= \dfrac{4G \cdot (1) \cdot (2) \cdot (1)}{(2^2 - 1^2)^2} \\ &= \boxed{5.93 \times 10^{-11} \dfrac{\text{N}}{\text{kg}}} \end{align*}$

We observe that $F_T > F_{T,M}$, therefore, the melon's tidal force on you is greater than the Moon's.

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