Content check — University Physics 1, Chapter 13: Gravitation

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 13: Gravitation

55 problems · 2 figures

Check Your Understanding 13.1 §13.1 source ↗

What happens to force and acceleration as the vehicles fall together? What will our estimate of the velocity at a collision higher or lower than the speed actually be? And finally, what would happen if the masses were not identical? Would the force on each be the same or different? How about their accelerations?

Answer

The force of gravity on each object increases with the square of the inverse distance as they fall together, and hence so does the acceleration. For example, if the distance is halved, the force and acceleration are quadrupled. Our average is accurate only for a linearly increasing acceleration, whereas the acceleration actually increases at a greater rate. So our calculated speed is too small. From Newton’s third law (action-reaction forces), the force of gravity between any two objects must be the same. But the accelerations will not be if they have different masses.

Check Your Understanding 13.2 §13.2 source ↗

How does your weight at the top of a tall building compare with that on the first floor? Do you think engineers need to take into account the change in the value of g when designing structural support for a very tall building?

Answer

The tallest buildings in the world are all less than 1 km. Since g is proportional to the distance squared from Earth’s center, a simple ratio shows that the change in g at 1 km above Earth’s surface is less than 0.0001%. There would be no need to consider this in structural design.

Check Your Understanding 13.3 §13.3 source ↗

Why not use the simpler expression

Δ U =

m g (y_{2} - y_{1})

? How significant would the error be? (Recall the previous result, in Example 13.4, that the value g at 400 km above the Earth is

8.67 {m/s}^{2}

Example 13.4 View it in the book ↗

Answer

The value of g drops by about 10% over this change in height. So $Δ U =$ $m g (y_{2} - y_{1})$ will give too large a value. If we use $g = 9.80 m/s$ , then we get

$Δ U =$ $m g (y_{2} - y_{1}) =$ $3.53 \times 10^{10} J$

which is about 6% greater than that found with the correct method.

Check Your Understanding 13.4 §13.3 source ↗

If we send a probe out of the solar system starting from Earth’s surface, do we only have to escape the Sun?

Answer

The probe must overcome both the gravitational pull of Earth and the Sun. In the second calculation of our example, we found the speed necessary to escape the Sun from a distance of Earth’s orbit, not from Earth itself. The proper way to find this value is to start with the energy equation, Equation 13.5, in which you would include a potential energy term for both Earth and the Sun.

Check Your Understanding 13.5 §13.3 source ↗

Assume you are in a spacecraft in orbit about the Sun at Earth’s orbit, but far away from Earth (so that it can be ignored). How could you redirect your tangential velocity to the radial direction such that you could then pass by Mars’s orbit? What would be required to change just the direction of the velocity?

Answer

You change the direction of your velocity with a force that is perpendicular to the velocity at all points. In effect, you must constantly adjust the thrusters, creating a centripetal force until your momentum changes from tangential to radial. A simple momentum vector diagram shows that the net change in momentum is

\sqrt{2}

times the magnitude of momentum itself. This turns out to be a very inefficient way to reach Mars. We discuss the most efficient way in Kepler’s Laws of Planetary Motion.

Check Your Understanding 13.6 §13.4 source ↗

By what factor must the radius change to reduce the orbital velocity of a satellite by one-half? By what factor would this change the period?

Answer

In Equation 13.7, the radius appears in the denominator inside the square root. So the radius must increase by a factor of 4, to decrease the orbital velocity by a factor of 2. The circumference of the orbit has also increased by this factor of 4, and so with half the orbital velocity, the period must be 8 times longer. That can also be seen directly from Equation 13.8.

Check Your Understanding 13.7 §13.4 source ↗

There is another consideration to this last calculation of

M_{E}

. We derived Equation 13.8 assuming that the satellite orbits around the center of the astronomical body at the same radius used in the expression for the gravitational force between them. What assumption is made to justify this? Earth is about 81 times more massive than the Moon. Does the Moon orbit about the exact center of Earth?

Answer

The assumption is that orbiting object is much less massive than the body it is orbiting. This is not really justified in the case of the Moon and Earth. Both Earth and the Moon orbit about their common center of mass. We tackle this issue in the next example.

Check Your Understanding 13.8 §13.4 source ↗

Galaxies are not single objects. How does the gravitational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?

Answer

The stars on the “inside” of each galaxy will be closer to the other galaxy and hence will feel a greater gravitational force than those on the outside. Consequently, they will have a greater acceleration. Even without this force difference, the inside stars would be orbiting at a smaller radius, and, hence, there would develop an elongation or stretching of each galaxy. The force difference only increases this effect.

Check Your Understanding 13.9 §13.5 source ↗

The nearly circular orbit of Saturn has an average radius of about 9.5 AU and has a period of 30 years, whereas Uranus averages about 19 AU and has a period of 84 years. Is this consistent with our results for Halley’s comet?

Answer

The semi-major axis for the highly elliptical orbit of Halley’s comet is 17.8 AU and is the average of the perihelion and aphelion. This lies between the 9.5 AU and 19 AU orbital radii for Saturn and Uranus, respectively. The radius for a circular orbit is the same as the semi-major axis, and since the period increases with an increase of the semi-major axis, the fact that Halley’s period is between the periods of Saturn and Uranus is expected.

Check Your Understanding 13.10 §13.6 source ↗

Earth exerts a tidal force on the Moon. Is it greater than, the same as, or less than that of the Moon on Earth? Be careful in your response, as tidal forces arise from the difference in gravitational forces between one side and the other. Look at the calculations we performed for the tidal force on Earth and consider the values that would change significantly for the Moon. The diameter of the Moon is one-fourth that of Earth. Tidal forces on the Moon are not easy to detect, since there is no liquid on the surface.

Answer

Consider the last equation above. The values of

r_{1}

and

r_{2}

remain nearly the same, but the diameter of the Moon,

(r_{2}^{} - r_{1}^{})

, is one-fourth that of Earth. So the tidal forces on the Moon are about one-fourth as great as on Earth.

Check Your Understanding 13.11 §13.7 source ↗

Consider the density required to make Earth a black hole compared to that required for the Sun. What conclusion can you draw from this comparison about what would be required to create a black hole? Would you expect the Universe to have many black holes with small mass?

Answer

Given the incredible density required to force an Earth-sized body to become a black hole, we do not expect to see such small black holes. Even a body with the mass of our Sun would have to be compressed by a factor of 80 beyond that of a neutron star. It is believed that stars of this size cannot become black holes. However, for stars with a few solar masses, it is believed that gravitational collapse at the end of a star’s life could form a black hole. As we will discuss later, it is now believed that black holes are common at the center of galaxies. These galactic black holes typically contain the mass of many millions of stars.

13-1 §13.1 source ↗

Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in science, and why was this action at a distance ultimately accepted?

Answer

The ultimate truth is experimental verification. Field theory was developed to help explain how force is exerted without objects being in contact for both gravity and electromagnetic forces that act at the speed of light. It has only been since the twentieth century that we have been able to measure that the force is not conveyed immediately.

13-3 §13.2 source ↗

Must engineers take Earth’s rotation into account when constructing very tall buildings at any location other than the equator or very near the poles?

Answer

The centripetal acceleration is not directed along the gravitational force and therefore the correct line of the building (i.e., the plumb bob line) is not directed towards the center of Earth. But engineers use either a plumb bob or a transit, both of which respond to both the direction of gravity and acceleration. No special consideration for their location on Earth need be made.

13-5 §13.3 source ↗

It was shown that the energy required to lift a satellite into a low Earth orbit (the change in potential energy) is only a small fraction of the kinetic energy needed to keep it in orbit. Is this true for larger orbits? Is there a trend to the ratio of kinetic energy to change in potential energy as the size of the orbit increases?

Answer

As we move to larger orbits, the change in potential energy increases, whereas the orbital velocity decreases. Hence, the ratio is highest near Earth’s surface (technically infinite if we orbit at Earth’s surface with no elevation change), moving to zero as we reach infinitely far away.

13-7 §13.4 source ↗

Many satellites are placed in geosynchronous orbits. What is special about these orbits? For a global communication network, how many of these satellites would be needed?

Answer

The period of the orbit must be 24 hours. But in addition, the satellite must be located in an equatorial orbit and orbiting in the same direction as Earth’s rotation. All three criteria must be met for the satellite to remain in one position relative to Earth’s surface. At least three satellites are needed, as two on opposite sides of Earth cannot communicate with each other. (This is not technically true, as a wavelength could be chosen that provides sufficient diffraction. But it would be totally impractical.)

13-9 §13.5 source ↗

In the diagram below for a satellite in an elliptical orbit about a much larger mass, indicate where its speed is the greatest and where it is the least. What conservation law dictates this behavior? Indicate the directions of the force, acceleration, and velocity at these points. Draw vectors for these same three quantities at the two points where the y-axis intersects (along the semi-minor axis) and from this determine whether the speed is increasing decreasing, or at a max/min.

A diagram showing an x y coordinate system and an ellipse, centered on the origin with foci on the x axis. The focus on the left is labeled f 1 and M. The focus on the right is labeled f 2. A location labeled as m is shown above f 2. The right triangle defined by f 1, f 2, and m is shown in red. The clockwise direction tangent to the ellipse is indicated by blue arrows.

Answer

The speed is greatest where the satellite is closest to the large mass and least where farther away—at the periapsis and apoapsis, respectively. It is conservation of angular momentum that governs this relationship. But it can also be gleaned from conservation of energy, the kinetic energy must be greatest where the gravitational potential energy is the least (most negative). The force, and hence acceleration, is always directed towards M in the diagram, and the velocity is always tangent to the path at all points. The acceleration vector has a tangential component along the direction of the velocity at the upper location on the y-axis; hence, the satellite is speeding up. Just the opposite is true at the lower position.

13-11 §13.7 source ↗

The principle of equivalence states that all experiments done in a lab in a uniform gravitational field cannot be distinguished from those done in a lab that is not in a gravitational field but is uniformly accelerating. For the latter case, consider what happens to a laser beam at some height shot perfectly horizontally to the floor, across the accelerating lab. (View this from a nonaccelerating frame outside the lab.) Relative to the height of the laser, where will the laser beam hit the far wall? What does this say about the effect of a gravitational field on light? Does the fact that light has no mass make any difference to the argument?

Answer

The laser beam will hit the far wall at a lower elevation than it left, as the floor is accelerating upward. Relative to the lab, the laser beam “falls.” So we would expect this to happen in a gravitational field. The mass of light, or even an object with mass, is not relevant.

13-13 §13.1 source ↗

Evaluate the magnitude of gravitational force between two 5-kg spherical steel balls separated by a center-to-center distance of 15 cm.

Answer

7.4 \times 10^{−8} N

13-15 §13.1 source ↗

Astrology makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational. (a) Calculate the gravitational force exerted on a 4.20-kg baby by a 100-kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some

6.29 \times 10^{11} m

away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

Answer

7.01 \times 10^{−7} N

; b. The mass of Jupiter is

\begin{matrix} m_{J} = 1.90 \times 10^{27} kg \\ F_{J} = 1.35 \times 10^{−6} N \\ \frac{F_{f}}{F_{J}} = 0.521 \end{matrix}

13-17 §13.1 source ↗

The International Space Station has a mass of approximately 370,000 kg. (a) What is the force on a 150-kg suited astronaut if she is 20 m from the center of mass of the station? (b) How accurate do you think your answer would be?

An image of the international space station is shown. — Figure 13.33 in the book ↗

Answer

9.25 \times 10^{−6} N

; b. Not very, as the ISS is not even symmetrical, much less spherically symmetrical.

13-19 §13.1 source ↗

(a) What was the acceleration of Earth caused by asteroid Toutatis (see previous problem) at its closest approach? (b) What was the acceleration of Toutatis at this point?

Answer

1.41 \times 10^{−15} {m/s}^{2}

; b.

1.69 \times 10^{−4} {m/s}^{2}

13-21 §13.2 source ↗

(a) What is the acceleration due to gravity on the surface of the Moon? (b) On the surface of Mars? The mass of Mars is

6.418 \times 10^{23} kg

and its radius is

3.38 \times 10^{6} m

Answer

1.62 {m/s}^{2}

; b.

3.75 {m/s}^{2}

13-23 §13.2 source ↗

The mass of a particle is 15 kg. (a) What is its weight on Earth? (b) What is its weight on the Moon? (c) What is its mass on the Moon? (d) What is its weight in outer space far from any celestial body? (e) What is its mass at this point?

Answer

a. 147 N; b. 25.5 N; c. 15 kg; d. 0; e. 15 kg

13-25 §13.2 source ↗

The mean diameter of the planet Saturn is

1.2 \times 10^{8} m

, and its mean mass density is

0.69 {g/cm}^{3}

. Find the acceleration due to gravity at Saturn’s surface.

Answer

12 {m/s}^{2}

13-27 §13.2 source ↗

The acceleration due to gravity on the surface of a planet is three times as large as it is on the surface of Earth. The mass density of the planet is known to be twice that of Earth. What is the radius of this planet in terms of Earth’s radius?

Answer

(3 / 2) R_{E}

13-29 §13.3 source ↗

Find the escape speed of a projectile from the surface of Mars.

Answer

5000 m/s

13-31 §13.3 source ↗

What is the escape speed of a satellite located at the Moon’s orbit about Earth? Assume the Moon is not nearby.

Answer

1440 m/s

13-33 §13.3 source ↗

An average-sized asteroid located

5.0 \times 10^{7} km

from Earth with mass

2.0 \times 10^{13} kg

is detected headed directly toward Earth with speed of 2.0 km/s. What will its speed be just before it hits our atmosphere? (You may ignore the size of the asteroid.)

Answer

11 km/s

13-35 §13.3 source ↗

(a) What is the change in energy of a 1000-kg payload taken from rest at the surface of Earth and placed at rest on the surface of the Moon? (b) What would be the answer if the payload were taken from the Moon’s surface to Earth? Is this a reasonable calculation of the energy needed to move a payload back and forth?

Answer

5.85 \times 10^{10} J

; b.

−5.85 \times 10^{10} J

; No. It assumes the kinetic energy is recoverable. This would not even be reasonable if we had an elevator between Earth and the Moon.

13-37 §13.4 source ↗

Two planets in circular orbits around a star have speeds of v and 2v. (a) What is the ratio of the orbital radii of the planets? (b) What is the ratio of their periods?

Answer

a. 0.25; b. 0.125

13-39 §13.4 source ↗

(a) What is the orbital radius of an Earth satellite having a period of 1.00 h? (b) What is unreasonable about this result?

Answer

5.08 \times 10^{3} km

; b. This less than the radius of Earth.

13-41 §13.4 source ↗

Find the mass of Jupiter based on the fact that Io, its innermost moon, has an average orbital radius of 421,700 km and a period of 1.77 days.

Answer

1.89 \times 10^{27} kg

13-43 §13.4 source ↗

(a) In order to keep a small satellite from drifting into a nearby asteroid, it is placed in orbit with a period of 3.02 hours and radius of 2.0 km. What is the mass of the asteroid? (b) Does this mass seem reasonable for the size of the orbit?

Answer

4.01 \times 10^{13} kg

; b. The satellite must be outside the radius of the asteroid, so it can’t be larger than this. If it were this size, then its density would be about

1200 {kg/m}^{3}

. This is just above that of water, so this seems quite reasonable.

13-45 §13.4 source ↗

The Sun orbits the Milky Way galaxy once each

2.60 \times 10^{8} years

, with a roughly circular orbit averaging a radius of

3.00 \times 10^{4}

light-years. (A light-year is the distance traveled by light in 1 year.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

Answer

1.66 \times 10^{−10} {m/s}^{2}

; Yes, the centripetal acceleration is so small it supports the contention that a nearly inertial frame of reference can be located at the Sun. b.

2.17 \times 10^{5} m / s

13-47 §13.5 source ↗

Calculate the mass of the Sun based on data for average Earth’s orbit and compare the value obtained with the Sun’s commonly listed value of

1.989 \times 10^{30} kg

Answer

1.98 \times 10^{30} kg

; The values are the same within 0.05%.

13-49 §13.5 source ↗

The “mean” orbital radius listed for astronomical objects orbiting the Sun is typically not an integrated average but is calculated such that it gives the correct period when applied to the equation for circular orbits. Given that, what is the mean orbital radius in terms of aphelion and perihelion?

Answer

Compare Equation 13.8 and Equation 13.11 to see that they differ only in that the circular radius, r, is replaced by the semi-major axis, a. Therefore, the mean radius is one-half the sum of the aphelion and perihelion, the same as the semi-major axis.

13-51 §13.5 source ↗

The perihelion of the comet Lagerkvist is 2.61 AU and it has a period of 7.36 years. Show that the aphelion for this comet is 4.95 AU.

Answer

The semi-major axis, 3.78 AU is found from the equation for the period. This is one-half the sum of the aphelion and perihelion, giving an aphelion distance of 4.95 AU.

13-53 §13.5 source ↗

Eros has an elliptical orbit about the Sun, with a perihelion distance of 1.13 AU and aphelion distance of 1.78 AU. What is the period of its orbit?

Answer

1.75 years

13-55 §13.6 source ↗

If the Sun were to collapse into a black hole, the point of no return for an investigator would be approximately 3 km from the center singularity. Would the investigator be able to survive visiting even 300 km from the center? Answer this by finding the difference in the gravitational attraction the black holes exerts on a 1.0-kg mass at the head and at the feet of the investigator.

Answer

19,800 N; this is clearly not survivable

13-57 §13.7 source ↗

What is the Schwarzschild radius for the black hole at the center of our galaxy if it has the mass of 4 million solar masses?

Answer

1.19 \times 10^{7} km

13-59 §13.AP source ↗

A neutron star is a cold, collapsed star with nuclear density. A particular neutron star has a mass twice that of our Sun with a radius of 12.0 km. (a) What would be the weight of a 100-kg astronaut on standing on its surface? (b) What does this tell us about landing on a neutron star?

Answer

1.85 \times 10^{14} N

; b. Don’t do it!

13-61 §13.AP source ↗

How far from the center of the Sun would the net gravitational force of Earth and the Sun on a spaceship be zero?

Answer

1.49 \times 10^{8} km

13-63 §13.AP source ↗

Suppose you can communicate with the inhabitants of a planet in another solar system. They tell you that on their planet, whose diameter and mass are

5.0 \times 10^{3} km

and

3.6 \times 10^{23} kg

, respectively, the record for the high jump is 2.0 m. Given that this record is close to 2.4 m on Earth, what would you conclude about your extraterrestrial friends’ jumping ability?

Answer

The value of g for this planet is 3.8 m/s², which is about one-fourth that of Earth. So they are weak high jumpers.

13-65 §13.AP source ↗

A body of mass 100 kg is weighed at the North Pole and at the equator with a spring scale. What is the scale reading at these two points? Assume that

g = 9.83 {m/s}^{2}

at the pole.

Answer

At the North Pole, 983 N; at the equator, 980 N

13-67 §13.AP source ↗

Consider the previous problem and include the fact that Earth has an orbital speed about the Sun of 29.8 km/s. (a) What speed relative to Earth would be needed and in what direction should you leave Earth? (b) What will be the shape of the trajectory?

Answer

a. The escape velocity is still 43.6 km/s. By launching from Earth in the direction of Earth’s tangential velocity, you need

43.4 -

29.8 =

13.8 km/s

relative to Earth. b. The total energy is zero and the trajectory is a parabola.

13-69 §13.AP source ↗

An asteroid has speed 15.5 km/s when it is located 2.00 AU from the sun. At its closest approach, it is 0.400 AU from the Sun. What is its speed at that point?

Answer

61.5 km/s

13-71 §13.AP source ↗

A satellite of mass 1000 kg is in circular orbit about Earth. The radius of the orbit of the satellite is equal to two times the radius of Earth. (a) How far away is the satellite? (b) Find the kinetic, potential, and total energies of the satellite.

Answer

1.3 \times 10^{7} m

; b.

1.56 \times 10^{10} J

;

− 3.12 \times 10^{10} J

;

−1.56 \times 10^{10} J

13-73 §13.AP source ↗

(a) Given the asteroid Vesta which has a diameter of 520 km and mass of

2.67 \times 10^{20} kg

, what would be the orbital period for a space probe in a circular orbit of 10.0 km from its surface? (b) Why is this calculation marginally useful at best?

Answer

6.24 \times 10^{3} s

or about 1.8 hours. This was using the 520 km average diameter. b. Vesta is clearly not very spherical, so you would need to be above the largest dimension, nearly 580 km. More importantly, the nonspherical nature would disturb the orbit very quickly, so this calculation would not be very accurate even for one orbit.

13-75 §13.AP source ↗

(a) Using the information in the previous problem, what velocity do you need to escape the Milky Way galaxy from our present position? (b) Would you need to accelerate a spaceship to this speed relative to Earth?

Answer

a. 323 km/s; b. No, you need only the difference between the solar system’s orbital speed and escape speed, so about

323 -

228 =

95 km/s

13-77 §13.AP source ↗

Show that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartesian coordinates, x and y, for the polar coordinates, r and

θ

, and showing that it has the general form for a parabola,

x = a y^{2} + b y + c

Answer

Setting

e = 1

, we have

\frac{α}{r} =

1 +

cos θ \to

α =

r +

r cos θ =

r +

x

; hence,

r^{2} =

x^{2} +

y^{2} =

{(α - x)}^{2}

. Expand and collect to show

x = \frac{1}{−2 α} y^{2} + \frac{α}{2}

13-79 §13.AP source ↗

Given the perihelion distance, p, and aphelion distance, q, for an elliptical orbit, show that the velocity at perihelion,

v_{p}

, is given by

v_{p} =

\sqrt{\frac{2 G M_{Sun}}{(q + p)} \frac{q}{p}}

. (Hint: Use conservation of angular momentum to relate

v_{p}

and

v_{q}

, and then substitute into the conservation of energy equation.)

Answer

Substitute directly into the energy equation using

p v_{p} = q v_{q}

from conservation of angular momentum, and solve for

v_{p}

13-81 §13.CP source ↗

A tunnel is dug through the center of a perfectly spherical and airless planet of radius R. Using the expression for g derived in Gravitation Near Earth’s Surface for a uniform density, show that a particle of mass m dropped in the tunnel will execute simple harmonic motion. Deduce the period of oscillation of m and show that it has the same period as an orbit at the surface.

Answer

g =

\frac{4}{3} G ρ π r \to

F =

m g =

[\frac{4}{3} G m ρ π] r

, and from

F = m \frac{d^{2} r}{d t^{2}}

, we get

\frac{d^{2} r}{d t^{2}} =

[\frac{4}{3} G ρ π] r

where the first term is

ω^{2}

. Then

T =

\frac{2 π}{ω} =

2 π \sqrt{\frac{3}{4 G ρ π}}

and if we substitute

ρ = \frac{M}{4 / 3 π R^{3}}

, we get the same expression as for the period of orbit R.

13-83 §13.CP source ↗

Show that the areal velocity for a circular orbit of radius r about a mass M is

\frac{Δ A}{Δ t} = \frac{1}{2} \sqrt{G M r}

. Does your expression give the correct value for Earth’s areal velocity about the Sun?

Answer

Using the mass of the Sun and Earth’s orbital radius, the equation gives

2.24 \times 10^{15} m^{2} /s

. The value of

π R_{ES}^{2} / (1 year)

gives the same value.

13-85 §13.CP source ↗

Show that for small changes in height h, such that

h < < R_{E}

, Equation 13.4 reduces to the expression

Δ U = m g h

Answer

$Δ U =$ $U_{f} -$ $U_{i} =$ $-$ $\frac{G M_{E} m}{r_{f}} +$ $\frac{G M_{E} m}{r_{i}} =$ $G M_{E} m (\frac{r_{f} - r_{i}}{r_{f} r_{i}})$ where $h = r_{f} - r_{i}$ . If $h < < R_{E}$ , then $r_{f} r_{i} \approx R_{E}^{2}$ , and upon substitution, we have

$Δ U =$ $G M_{E} m (\frac{h}{R_{E}^{2}}) =$ $m (\frac{G M_{E}}{R_{E}^{2}}) h$ where we recognize the expression with the parenthesis as the definition of g.

13-87 §13.CP source ↗

(a) Show that tidal force on a small object of mass m, defined as the difference in the gravitational force that would be exerted on m at a distance at the near and the far side of the object, due to the gravitation at a distance R from M, is given by

F_{tidal} =

\frac{2 G M m}{R^{3}} Δ r

where

Δ r

is the distance between the near and far side and

Δ r < < R

. (b) Assume you are falling feet first into the black hole at the center of our galaxy. It has mass of 4 million solar masses. What would be the difference between the force at your head and your feet at the Schwarzschild radius (event horizon)? Assume your feet and head each have mass 5.0 kg and are 2.0 m apart. Would you survive passing through the event horizon?

Answer

a. Find the difference in force,

F_{tidal} =

=

\frac{2 G M m}{R^{3}} Δ r

;
b. For the case given, using the Schwarzschild radius from a previous problem, we have a tidal force of

9.5 \times 10^{−3} N

. This won’t even be noticed!

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Problems and answers are adapted from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.