Unit 9: Gravity

There is no change in the compression of the springs inside the bathroom scale in either scenario.

While moving at constant velocity (either upward or downward), the elevator system experiences no acceleration. This means that while standing on a bathroom scale inside the elevator, your acceleration (as well as the net force on your body) is equal to zero. Therefore, you conclude that the downward force of gravity (or your weight) is balanced by the upward normal (or support) force exerted by the bathroom scale on your feet. Since the scale is sensitive to the normal force, the measured weight would be equal to your true weight.

Tides depend more on the difference in strength of gravitational pull.

The gravitational pull of the Sun on Earth is 180 times greater than the gravitational pull of the Moon on Earth. Nevertheless, the ocean tides are caused mostly by the gravitational influence of the Moon. This is because the tidal effect depends on the difference in gravitational pull on opposite sides of Earth. Such difference in the lunar field is 2.2 times greater than the corresponding difference in the solar field.

9.8 N

To calculate the force of gravity on a body of mass $m = 1\,\text{kg}$ at Earth’s surface, substitute into the universal law of gravity, where $M_\text{E} = 6.0\times 10^{24}\,\text{kg}$ and $R_\text{E} = 6.4 \times 10^6\,\text{m}$, as follows: \begin{align} F_g &= G\, \frac{m M_\text{E}}{R_\text{E}^2} \nonumber\\[6pt] &= (6.67 \times 10^{-11}\, \text{N}\cdot\text{m}^2/\text{kg}^2)\times \frac{(1\,\text{kg}) (6.0\times 10^{24}\,\text{kg})}{(6.4 \times 10^6\,\text{m})^2} \nonumber\\[6pt] &= 9.8\,\text{N} \end{align} As expected, this is equal to the weight of a 1-kg object near Earth’s surface, where you have \begin{align} \text{Weight} &= m g \nonumber\\[6pt] &= (1\,\text{kg}) (9.8\,\text{m}/\text{s}^2) \nonumber\\[6pt] &= 9.8\,\text{N} \end{align} Note that in other questions in this course, the approximate value of $10\,\text{m}/\text{s}^2$ is used for the acceleration due to gravity.

$2.45\,\text{N}$

You can calculate the force of gravity on a 1-kg mass located at two Earth radii from Earth’s center by using a calculation similar to the one in the answer to the previous question: \begin{align} F_g &= G\, \frac{m M_\text{E}}{(2 R_\text{E})^2} \nonumber\\[6pt] &= \frac{1}{4} \left(G\, \frac{m M_\text{E}}{R_\text{E}^2}\right) \nonumber\\[6pt] &= \frac{9.8\,\text{N}}{4} \nonumber\\[6pt] &= 2.45\,\text{N} \end{align} Note that this value is one-quarter of the value you obtained in the answer to the previous question. This is because the force of gravity decreases by four times when the distance doubles.

Starting with the gravitational force equation \begin{equation} F_1 = G\, \frac{m_1\, m_2}{r^2} \end{equation} when you double the masses of the two objects ($m_1$ and $m_2$) and also the distance ($r$) between them, you can write \begin{align} F_2 &= G\, \frac{(2 m_1) (2 m_2)}{(2r)^2} \nonumber\\[6pt] &= G\, \frac{4\, m_1\, m_2}{4\, r^2} \nonumber\\[6pt] &= G\, \frac{m_1\, m_2}{r^2} \nonumber\\[6pt] &= F_1 \end{align} You see that doubling each mass contributes a factor of 2, while doubling the distance contributes a factor of 1⁄4.

1. $\text{B}$, $\text{A} = \text{C}$, then $\text{D}$
2. $\text{D}$, $\text{A} = \text{C}$, then $\text{B}$

1. The gravitational force below the ground pulls your body toward Earth’s center, and the longer it pulls on you, the more speed you gain. Hence, the greatest speed is attained at position B (see Figure Q9.47), Earth’s center. After passing position B, gravity acts to pull you back toward Earth’s center, thus slowing you down until you stop completely when you reach position D. So the further away you are from the center, the slower your speed is.
2. The gravitational force inside Earth decreases as you approach the center (see Figure 9.24 in the eText). Therefore, your acceleration is greatest near Earth’s surface and decreases as you fall deeper into the hole, until it becomes zero at the center.

The acceleration due to gravity is less atop Mt. Everest than at sea level. The force of gravity (and consequently the acceleration due to gravity) decreases with increasing distance from Earth’s center. Since the peak of Mt. Everest is nearly 9 km above sea level, the acceleration due to gravity decreases as you climb the mountain. When you calculate the acceleration due to gravity ($g$), you find that it drops by approximately 0.3% at the top of the mountain.

Apparent weight is equal to the support force, which is absent during free fall, and equal to $mg$ while an object is falling at terminal velocity.

A person falling freely (in the absence of air resistance) does not experience any upward force; therefore, their body is not pressed against any object to create a sense of weight. As a result, the person feels weightless in this situation. However, when a person is falling with constant terminal velocity, the net force on the person is equal to zero. In this case, air drag provides an upward support force equal to the person’s weight and creates a sense of weight.

Yes, in both situations.

Gravitational forces always exist between objects regardless of whether they are in motion or at rest. Therefore, the force of gravity between your body (in jetsuit) and Earth and between an astronaut (in a space shuttle) and Earth cannot be turned off or shielded by any material or vacuum.

Because the lunar day is longer than the solar day.

While Earth spins about its axis, the Moon advances in its orbit around Earth. Therefore, a specific location on Earth directly facing the Moon will take a full lunar day (24 hours and 50 minutes) for the next alignment with the Moon (see Video Q9.88). As a result, high ocean tides occur 12 hours and 25 minutes apart.

It is true that the atmosphere is held around Earth by gravity and that air density decreases with altitude and practically disappears after a few hundred kilometers. However, the force of gravity does not vanish past the atmosphere. The gravitational force equation shows that the force exerted by Earth on a space shuttle diminishes as the shuttle travels farther from Earth, but $F$ never becomes zero. For example, at an altitude of 500 km, the force of gravity on the space shuttle is only 14% less than what it is on the ground a the Kennedy Space Center.

Note that an orbiting space shuttle moves around Earth along a circular orbit, where the force of gravity provides the required centripetal force.

This location is much closer to the Moon.

Earth is 81 times more massive than the Moon. Therefore, a space pod located midway between Earth and the Moon experiences a pull toward Earth that is 81 times stronger than that toward the Moon.

The force of gravity, however, is also inversely proportional to the square of the distance. So if the space pod moves closer to the Moon such that the distance to Earth ($d_\text{E}$) is nine times the distance to the Moon ($d_\text{M}$), the pod will experience equal forces from the two bodies, resulting in a zero net force (see Figure Q9.100). In this case, the factor of 81 due to the larger mass is canceled by a factor of $1/9^2 = 1/81$ due to the greater distance.

Mathematically speaking, when $M_\text{E} = 81\, M_\text{M}$ and $d_\text{E} = 9\, d_\text{M}$, the force of gravity on the pod (of mass $m$) due to the Moon is given by \begin{equation} F_\text{M} = G\, \frac{m\, M_\text{M}}{d_\text{M}^2} \end{equation} In this case, the gravitational force applied by Earth is \begin{align} F_\text{E} &= G\, \frac{m\, M_\text{E}}{d_\text{E}^2} \nonumber\\[6pt] &= G\,\frac{m\, (81\, M_\text{M})}{(9\, d_\text{M})^2} \nonumber\\[6pt] &= G\,\frac{m\, (81\, M_\text{M})}{81\, d_\text{M}^2} \nonumber\\[6pt] &= G\,\frac{m\, M_\text{M}}{d_\text{M}^2} \nonumber\\[6pt] &= F_\text{M} \end{align}

The increase in the body’s weight due to Jupiter’s mass is partially offset by Jupiter’s size.

The force of gravity exerted by a planet on a body is proportional to its mass and inversely proportional to the square of the distance to the planet’s center. If Jupiter were the same size as Earth, a body near the surface of Jupiter would weigh 300 times more than on Earth’s surface. This is because Jupiter is 300 times more massive than Earth. The radius of Jupiter, however, is roughly 10 times that of Earth (see Figure Q9.103). This reduces the gravitational attraction by a factor of $10^2 = 100$ on the surface of the giant planet. When both factors are combined, the net increase in the body’s weight is only by a factor of $300\times 1/100 = 3$.

All objects on the Earth are in a state of continuous free fall around the Sun.

While orbiting the Sun, the entire Earth system experiences an acceleration due to solar gravity that is approximately equal to $0.006\,\text{m}/\text{s}^2$. This is also equal to the centripetal acceleration directed toward the Sun, which every object on Earth experiences at all times.

So when you stand on a bathroom scale, both you and the scale rotate around the Sun with the same speed and acceleration. As a result, the Sun’s gravitational attraction does not contribute to the contact force between your feet and the scale. In other words, the rotation around the Sun does not cause your feet to press harder (or less hard) on the scale, which in turn does not change the support force it exerts on your feet. Therefore, you should not feel any change in your weight between noon and midnight due to the position of the Sun.

The Moon does rotate about its axis, and its spin about its axis is synchronized with its rotation around Earth. This means that the time it takes the Moon to complete one rotation about its axis is equal to the period of its complete revolution around Earth. In Video Q9.108 (not drawn to scale), you can see how the Moon spins slowly about its axis such that only one side faces Earth all the time. Note that if the Moon did not rotate about its axis, you would see all parts of the Moon’s surface during a lunar month.

The force of gravity on a body of mass $m$ is given by the equation

\begin{equation} F_g = G\, \frac{m\, M_\text{E}}{r^2} \label{u09_AQA_Fg} \end{equation}

where $M_\text{E}$ is the mass of Earth and $r$ is the distance between the body and the center of Earth. According to Newton’s second law, the acceleration of a body under the influence of gravity only is given by

\begin{equation} a = \frac{F_g}{m} \label{u09_AQA_aFm} \end{equation}

By substituting from Equation 9.16 into Equation 9.17, you have

\begin{align} a &= \left(G\, \frac{m\, M_\text{E}}{d^2}\right) \times \frac{1}{m} \nonumber\\[6pt] &= G\, \frac{M_\text{E}}{d^2} \label{u09_AQA_aQMEd}\textrm{.} \end{align}

From Equation 9.18, you can see that the acceleration of a freely falling body is independent of its mass.