HW Pract 07 - Gravitation practice materials PDF

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Gravity practice problems that are descriptive of many things including orbits of planets...

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AP Physics 1

Name: __________________ Date: ____________ Prd: ___

HW Practice 07 – Gravitation Questions 13.

Does an apple exert a gravitational force on the Earth? If so, how large a force? Consider an apple (a) attached to a tree and (b) falling.

14.

Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?

18.

When will your apparent weight be the greatest, as measured by a scale in a moving elevator: when the elevator (a) accelerates downward, (b) accelerates upward, (c) is in free fall, or (d) moves upward at constant speed? (e) In which case would your apparent weight be the least? (f) When would it be the same as when you are on the ground? Explain.

19.

The source of the Mississippi River is closer to the center of the Earth than is its outlet in Louisiana (because the Earth is fatter at the equator than at the poles). Explain how the Mississippi can flow “uphill.”

21.

Is the centripetal acceleration of Mars in its orbit around the Sun larger or smaller than the centripetal acceleration of the Earth? Explain.

MisConceptual Questions 7.

The Moon does not crash into the Earth because:

(a) (b) (c) (

the net force on it is zero. it is beyond the main pull of the Earth’s gravity. it is being pulled by the Sun as well as by the Earth. it is freely falling but it has a high tangential velocity.

8.

Which pulls harder gravitationally, the Earth on the Moon, or the Moon on the Earth? Which accelerates more?

(a) (b) (c) (d) (e)

The Earth on the Moon; the Earth. The Earth on the Moon; the Moon. The Moon on the Earth; the Earth. The Moon on the Earth; the Moon. Both the same; the Earth. Both the same; the Moon.

Mass of earth is 5.98 x 10^24kg radius of earth is 6.38 x 10^6 m

9.

In the International Space Station which orbits Earth, astronauts experience apparent weightlessness because

(a) (b) (d) (e)

the station is so far away from the center of the Earth. the station is kept in orbit by a centrifugal force that counteracts the Earth’s gravity. the astronauts and the station are in free fall towards the center of the Earth. there is no gravity in space. the station’s high speed nullifies the effects of gravity.

10

Two satellites orbit the Earth in circular orbits of the same radius. One satellite is twice as massive as the other. Which statement is true about the speeds of these satellites?

(a) (c) (d)

The heavier satellite moves twice as fast as the lighter one. The two satellites have the same speed. The lighter satellite moves twice as fast as the heavier one. The ratio of their speeds depends on the orbital radius.

Problems 5–5 and 5–6 Law of Universal Gravitation 28.

(I) Calculate the force of Earth’s gravity on a spacecraft 2.00 Earth radii above the Earth’s surface if its mass is 1850 kg.

29.

2010 N

(I) At the surface of a certain planet, the gravitational acceleration g has a magnitude of 12.0 m/s2. A 24.0-kg brass ball is transported to this planet. What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?

30.

235

288

(II) At what distance from the Earth will a spacecraft traveling directly from the Earth to the Moon experience zero net force because the Earth and Moon pull in opposite directions with equal force?

32.

Text

(II) A hypothetical planet has a radius 2.0 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?

37.

(II) A hypothetical planet has a mass 2.80 times that of Earth, but has the same radius. What is g near its surface?

38.

(II) If you doubled the mass and tripled the radius of a planet, by what factor would g at its surface change?

5–7 Satellites and Weightlessness 46.

(I) Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4800 km.

49.

(II) Calculate the period of a satellite orbiting the Moon, 95 km above the Moon’s surface. Ignore effects of the Earth. The radius of the Moon is 1740 km.

50.

(II) Two satellites orbit Earth at altitudes of 7500 km and 15,000 km above the Earth’s surface. Which satellite is faster, and by what factor?

HW Practice 07 – Gravitation Key Responses to Questions 13.

Whether the apple is (a) attached to a tree or (b) falling, it exerts a gravitational force on the Earth equal to the force the Earth exerts on it, which is the weight of the apple (Newton’s third law). That force is independent of the motion of the apple.

14.

Since the Earth’s mass is much greater than the Moon’s mass, the point at which the net gravitational pull on the spaceship is zero is closer to the Moon. It is shown in Problem 30 that this occurs at about 90% of the way from the Earth to the Moon. So, a spaceship traveling from the Earth toward the Moon must therefore use fuel to overcome the net pull backward for 90% of the trip. Once it passes that point, the Moon will exert a stronger pull than the Earth and accelerate the spacecraft toward the Moon. However, when the spaceship is returning to the Earth, it reaches the zero point at only 10% of the way from the Moon to the Earth. Therefore, for most of the trip toward the Earth, the spacecraft is “helped” by the net gravitational pull in the direction of travel, so less fuel is used.

18.

Your apparent weight will be greatest in case (b), when the elevator is accelerating upward. The scale reading (your apparent weight) indicates your force on the scale, which, by Newton’s third law, is the same as the normal force of the scale on you. If the elevator is accelerating upward, then the net force must be upward, so the normal force (up) must be greater than your actual weight (down). When in an elevator accelerating upward, you “feel heavy.” Your apparent weight will be least in case (c), when the elevator is in free fall. In this situation your apparent weight is zero since you and the elevator are both accelerating downward at the same rate and the normal force is zero. Your apparent weight will be the same as when you are on the ground in case (d), when the elevator is moving upward at a constant speed. If the velocity is constant, acceleration is zero and N = mg. (Note that it doesn’t matter if the elevator is moving up or down or even at rest, as long as the velocity is constant.)

19.

If the Earth were a perfect, nonrotating sphere, then the gravitational force on each droplet of water in the Mississippi would be the same at the headwaters and at the outlet, and the river wouldn’t flow. Since the Earth is rotating, the droplets of water experience a centripetal force provided by a part of the component of the gravitational force perpendicular to the Earth’s axis of rotation. The centripetal force is smaller for the headwaters, which are closer to the North Pole, than for the outlet, which is closer to the equator. Since the centripetal force is equal to mg – N (apparent weight) for each droplet, N is smaller at the outlet, and the river will flow. This effect is large enough to overcome smaller effects on the flow of water due to the bulge of the Earth near the equator.

21.

The centripetal acceleration of Mars in its orbit around the Sun is smaller than that of the Earth. For both planets, the centripetal force is provided by gravity, so the centripetal acceleration is inversely proportional to the square of the distance from the planet to the Sun:

mpu 2 r

=

Gms mp r

2

so

aR =

u2 r

=

Gms r2

Since Mars is at a greater distance from the Sun than is Earth, it has a smaller centripetal acceleration. Note that the mass of the planet does not appear in the equation for the centripetal acceleration.

Responses to MisConceptual Questions 7.

(d)

If the net force on the Moon were zero (answer (a)), the Moon would move in a straight line and not orbit about the Earth. Gravity pulls the Moon away from the straight-line motion. The large tangential velocity is what keeps the Moon from crashing into the Earth. The gravitational force of the Sun also acts on the Moon, but this force causes the Earth and Moon to orbit the Sun.

8.

(f)

A common misconception is that since the Earth is more massive than the Moon, it must exert more force. However, the force is an interaction between the Earth and Moon, so by Newton’s third law, the forces must be equal. Since the Moon is less massive than the Earth and the forces are equal, the Moon has the greater acceleration.

9.

(c)

The nonzero gravitational force on the ISS is responsible for it orbiting the Earth instead of moving is a straight line through space. Astronauts aboard the ISS experience the same centripetal acceleration (free fall toward the Earth) as the station and as a result do not experience a normal force (apparent weightlessness).

10.

(b)

A common misconception is that the mass of an object affects its orbital speed. However, as with all objects in free fall, when calculating the acceleration the object’s mass is divided out of the gravitational force. All objects at the same radial distance from the Earth experience the same centripetal acceleration, and by Eq. 5–1 they have the same orbital speed.

Solutions to Problems 5–5 and 5–6 Law of Universal Gravitation 28.

The spacecraft is at 3.00 Earth radii from the center of the Earth, or three times as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance, the force of gravity on the spacecraft will be one-ninth of its weight at the Earth’s surface. (1850 kg)(9 .80 m/s 2) FG = 19 mgEarth's = = 2010 N 9 surface

29.

(a) (b)

Mass is independent of location, so the mass of the ball is 24.0 kg on both the Earth and the planet. The weight is found by using W = mg . W Earth = mg Earth = (24.0 kg)(9.80 m/s2 ) = 235 N W Planet = mg Planet = (24.0 kg)(12.0 m/s 2 ) = 288 N

30.

For the net force to be zero means that the gravitational force on the spacecraft due to the Earth must be the same as that due to the Moon. Write d the gravitational forces on the Moon spacecraft, equate them, and solve for Earth the distance x. We measure from the center of the bodies. spacecraft x d–x FEarth-

=G

M Earth mspacecraft x

spacecraft

G

M Earth mspacecraft

=G

x2

x =d

M Earth

(

M Moon + M Earth

)

2

FMoon

;

=G

MMoon mspacecraft (d - x ) 2

spacecraft

M Moon mspacecraft (d - x)

= (3.84 ´108 m)

x2

®

2

M Earth

=

(d - x ) 2 M Moon

®

5.97 ´10 24 kg

(

7 .35 ´10 22 kg + 5 .97 ´10 24 kg

x M Earth

)

=

d-x M Moon

= 3 .46 ´108 m

This is only about 22 Moon radii away from the Moon. Alternatively, it is about 90% of the distance from the center of the Earth to the center of the Moon. 32.

The acceleration due to gravity at any location on or above the surface of a planet is given by g planet = GM planet /r 2, where r is the distance from the center of the planet to the location in question.

gplanet = G 37.

r

2

=G

M Earth (2 .0 REarth

)2

=

1 (2 .0) 2

G

M Earth 2 REarth

=

1 9.80 m/s 2 = 2.5 m/s 2 gEarth = 4 .0 4 .0

The acceleration due to gravity at any location at or above the surface of a planet is given by g planet = GM Planet /r 2, where r is the distance from the center of the planet to the location in question.

gplanet = G 38.

M Planet

M Planet r

2

=G

2.80M Earth 2 REarth

æ M = 2 .80 çG Earth ç R2 Earth è

ö 2 2 ÷÷ = 2 .80gEarth = 2 .80(9. 80 m/s ) = 27. 4 m/s ø

The acceleration due to gravity is determined by the mass of the Earth and the radius of the Earth. g0 =

GM 0 r02

gnew =

GM new G 2M 0 2 GM 0 2 = = = 9 g0 2 rnew (3 r0 )2 9 r02

So g is multiplied by a factor of 2/9 .

5–7 Satellites and Weightlessness 46.

The speed of a satellite in a circular orbit around a body is given in Example 5–12 as uorbit = GM body /r , where r is the distance from the satellite to the center of the body.

u= G

Mbody

= G

r = 5. 97´ 103 m/s

MEarth REarth + 4. 8´ 106 m

= (6.67 ´ 10-11 N × m2 /kg2 )

(5.98 ´ 10 24 kg) (6. 38 ´ 106 m + 4. 8´ 106 m)

49.

The speed of an object in a circular orbit of radius r around mass M is given in Example 5–12 by u = GM /r and is also given by u = 2p r /T , where T is the period of the orbiting object. Equate the two expressions for the speed and solve for T. G

M 2p r = r T

T = 2p

50.

®

(1 .74 ´ 106 m + 9. 5´ 104 m)3 r3 = 2p = 7 .05 ´ 103 s » 118 min GM (6. 67´ 10 -11 N× m2 /kg2 )(7. 35´ 1022 m)

The speed of a satellite in circular orbit around the Earth is shown in Example 5–12 to be M u orbit = G Earth . Thus the velocity is inversely related to the radius, so the closer satellite will be orbiting r faster. GM Earth rclose r R uclose 6. 38´ 106 m + 1. 5´ 107 m + 1. 5´ 107 m = = far = Earth = = 1.24 6 rclose u far 6. 38´ 106 m + 7. 5 ´ 106 m GM Earth REarth + 7. 5´ 10 m rfar So the close satellite is moving 1.2 times faster than the far satellite....