Circular Motion and Gravitation

vr

T

2

v = 2π r f

fT

1

vr

T

2

v = 2π r f

fT

1

r

GMv E

2

tanv

gR

2

tanv

gR

Fmv

rc 2

Fc = mac

r

vac

2

2

24

t

rac

UNIFORM CIRCULAR MOTION

Uniform circular motion is motion in which there is no change in speed, only a change in direction.

The period T is the time for one complete revolution. So the linear speed can be found by dividing the period into the circumference:

vr

T

2Units: m/s

fT

1

Another useful parameter in engineering problems is the rotational speed, expressed in revolutions per minute (rpm), revolutions per second (rev/s) or Hz (s-1). This quantity is called the frequency f of rotation and is given by the reciprocal of the period.

Units: s-1 = Hz

Speed/Velocity in a CircleConsider an object moving in a circle around a specific origin. The DISTANCE the object covers in ONE REVOLUTION is called the CIRCUMFERENCE. The TIME that it takes to cover this distance is called the PERIOD.

T

r

T

dscircle

2

Speed is the MAGNITUDE of the velocity. And while the speed may be constant, the VELOCITY is NOT. Since velocity is a vector with BOTH magnitude AND direction, we see that the direction o the velocity is ALWAYS changing.

We call this velocity, TANGENTIAL velocity as its direction is draw TANGENT to the circle.

The term centripetal means that the acceleration a is always directed toward the center. The velocity and the acceleration are not in the same direction; v points in the direction of motion which is tangential to the circle. v and a are perpendicular at every point.

v and a are perpendicular at every point.

CENTRIPETAL ACCELERATION

An object experiencing uniform circular motion is continually accelerating. The position and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure.

When the particle is at point A, its velocity is represented by vector v1. After a time

interval t, its velocity is represented by the vector v2.

The acceleration is given by:

t

vv

t

va

12

radius

chord

R

s

v

v

as Δt becomes smaller and smaller, the chord length becomes equal to the arc length s = v Δt

R

tv

v

v

R

v

t

v 2

2

c

va

R

This acceleration is called the centripetal, and it points toward the center of the circle.

Units: m/s2

CENTRIPETAL ACCELERATION

An object experiencing uniform circular motion is continually accelerating. The direction and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. The vectors are the same size because the velocity is constant but the changing direction means acceleration is occurring.

r

vac

2

To calculate the centripetal acceleration, we will use the linear velocity and the radius of the circle

Or substituting for v We get

2

24

t

rac

Example 2: A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration.

2

24

t

r

2

2

)2(

)6.0(4

s

m

= 5.92 m/s2

r

vac

2

r = 0.6 mT = 2 s

Centripetal Acceleration

metersin length arc

sr

s

v

v

r

vt

vtsv

v

r

s

Suppose we had a circle with angle, between 2 radaii. You may recall:

vo

v

v

vov

onacceleratilcentripetaa

at

v

r

v

c

c

2

Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circle

Drawing the Directions correctly

So for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and the acceleration would be

drawn TOWARDS the CENTER.

To find the MAGNITUDES of each we have:

r

va

T

rv cc

22

Circular Motion and N.S.L

Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true:

ForcelCentripetaFr

mvFF

r

vamaF

c

cNET

cNET

2

2

Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.

NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.

Uniform circular motionUniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction.

Constant force toward the center.

Constant velocity tangent to path.

vFc

Uniform Circular Motion (Cont.)The question of an The question of an outwardoutward force can be force can be resolved by asking what happens when the resolved by asking what happens when the string breaks! string breaks!

When central force is removed, ball continues in straight line.

vBall moves tangent Ball moves tangent to path, to path, NOTNOT outward as might outward as might be expected.be expected.

Centripetal forceCentripetal force is needed to change is needed to change direction.direction.

Examples

T

rvc

2

smvc /26.4)4*28(.

)76(.2

The blade of a windshield wiper moves through an angle of 90 degrees in 0.28seconds. The tip of the blade moves on the arc of a circle that has a radius of 0.76m. What is the magnitude of the centripetal acceleration of the tip of the blade?

222

/92.2376.0

)26.4(sm

r

vac

4.11 A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its linear speed and its centripetal acceleration.

m = 2 kgr = 2 mf = 3 rev/s

3

11

fT = 0.33 s

= 38.07 m/s

r

vac

2

= 725 m/s2

UCM

2 rv

T

2 (2)

0.33

2(38.07)

2

4.12 A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration.

r = 0.6 mT = 2 s

2

2

2

c

rv T

ar r

2

2

4 r

T

2

2

4 (0.6)

2

= 5.92 m/s2

UCM

CENTRIPETAL FORCEThe inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's Second Law, the centripetal force is given by:

c cF ma2

c

mvF

r

CENTRIPETAL FORCE The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's Second Law, the centripetal force is given by:

Fmv

rc 2

Fc = macOr

We can see that the force must be inward by thinking about a ball on a string:

The centripetal force is not a 'special' kind of force. The centripetal force is provided by the force that keeps the object in a circle, this is called the centripetal force requirement.

4.13 a. A car makes a turn, what force is required to keep it in circular motion?

As a car makes a turn, the force of friction acting upon the tires of the car provide the centripetal force required for circular motion.

Example 3: A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/sa. How much centripetal force is required?

b. Where does this force come from?

Force of friction between tires and road.

m = 1000 kgr = 30 mv = 9 m/s

r

mvFc

2

m

smkg

30

)/9)(1000( 2

= 2700 N

Figure 6.32• Two paths around a race track curve

are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed.

Examples

rg

v

r

mvmg

r

mvF

FF

N

cf

2

2

2

What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenthe penny is placed at the outer edge of the record?

mg

FN

Ff

Top view

Side view

187.0)8.9)(15.0(

)524.0(

/524.080.1

)15.0(22

sec80.1555.0

sec1

sec555.0sec60

min1*

min3.33

22

rg

v

smT

rv

Trevrev

revrev

c

 b. A bucket of water is tied to a string and spun in a circle, what force is required to keep it in circular motion?

c. The moon orbits the Earth, what force is required to keep it in circular motion?

As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

Without a centripetal force, an object in motion continues

along a straight-line path.

With a centripetal force, an object in motion will be

accelerated and change its direction.

                           

The Forbidden F-Word

When the subject of circular motion is discussed, it is not uncommon to hear mention of the word "centrifugal."

Centrifugal, not to be confused with centripetal, means away from the center or outward. The use of this word combined with the common sensation of an outward force when experiencing circular motion, often creates or reinforces a deadly student misconception.

The Forbidden F-Word

The deadly misconception, is the notion that objects in circular motion are experiencing an outward force.

"After all," a well-meaning student may think, "I can recall vividly the sensation of being thrown outward away from the center of the circle on that roller coaster ride. Therefore, circular motion must be characterized by an outward force."

Let’s imagine that you are riding in Granny’s car going around a curve.  Sitting on your dashboard is a cassette tape.  As you go around the curve, the tape moves to outside edge of the car.  Because you don't want to blame it on ghosts, you say “centrifugal force pushed the tape across the dashboard.”

Wwrroonngg!! 

Centrifugal Force is often used to describe why mud gets spun off a spinning tire, or water gets pushed out of the clothes during the spin dry cycle of your washer.  It is also used to describe why we tend to slide to the outer side of a car going around a curve. 

The animation shows both views at the same time.  The top window shows you the passenger's view of the car and the tape, while the shows you the bird's eye view.

The tape on the slippery dashboard does not have enough friction to act as a centripetal force, so in the absence of a centripetal force the tape follows straight line motion.  If the car you are riding in has the windows rolled down, then the tape will leave the car (or does the car leave the tape?) as it follows its straight line path.  If the windows are rolled up, then the window will deliver a centripetal force to the tape and keep it in a circular path.

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.

If the centripetal force vanishes, the object flies off tangent to the circle.

Spin Cycle on a Washer

How is the water removed How is the water removed from clothes during the spin from clothes during the spin cycle of a washer?cycle of a washer?

Think carefully before answering . . . Does the Think carefully before answering . . . Does the centripetal force throw water off the clothes?centripetal force throw water off the clothes?

NO.NO. Actually, it is the Actually, it is the LACKLACK of a force that of a force that allows the water to leave the clothes through allows the water to leave the clothes through holes in the circular wall of the rotating washer.holes in the circular wall of the rotating washer.

NO.NO. Actually, it is the Actually, it is the LACKLACK of a force that of a force that allows the water to leave the clothes through allows the water to leave the clothes through holes in the circular wall of the rotating washer.holes in the circular wall of the rotating washer.

The Conical PendulumA conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.

h

T

L

R mg

T

T sin

T cos

Note: The inward component of Note: The inward component of tension tension T sin T sin gives the needed gives the needed

central force.central force.

Angle and velocity v:

h

T

L

R mg

T

T sin

T cos

T cos = mg

mv2

RT sin Solve two

equations to find angle

tan = v 2

gR

Example 5: A 2-kg mass swings in a horizontal circle at the end of a cord of

length 10 m. What is the constant speed of the mass if the rope makes an angle of

300 with the vertical?

R = L sin 300 = (10 m)(0.5)

R = 5 m

1. Draw & label 1. Draw & label sketch.sketch.2. Recall formula for 2. Recall formula for pendulum.pendulum.

2

tanv

gR

2

tanv

gR Find: Find: v = ?v = ?

3. To use this formula, we need to find R 3. To use this formula, we need to find R = ?= ?

hT

L

R

Example 5(Cont.): Find v for = 300

R = R = 5 5 mm

v = 5.32 m/s

v = 5.32 m/s

g = g = 9.8 9.8 m/m/ss22

Solve for Solve for v = ?v = ?

2

tanv

gR

2

tanv

gR

4. Use given info to find 4. Use given info to find the velocity at the velocity at 303000..

2 tanv gR tanv gR

2 0(9.8 m/s )(5 m) tan 30v

hT

L

R

R = R = 5 m5 m

Example 5b: Now find the tension T in the cord if m = 2 kg, = 300, and L = 10

m.

h

T

L

R mg

T

T sin

T cos

Fy = 0: T cos - mg = 0; T cos = mg

T = =mg

cos

(2 kg)(9.8 m/s2)

cos 300

T = 22.6 NT = 22.6 N

2 kg

Example 6: Find the centripetal force Fc for the previous example.

h

T

L

R mg

T

T sin

T cos

m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N

Fc = 11.3 NFc = 11.3 N

2 kg

Fc = mv2

R

or Fc = T sin 300

Fc

= 30= 3000

4.14 A 75 g toy airplane is fastened to one end of a 44 cm string, and the other end is held fixed at the ceiling The plane whirls in a horizontal circle. Find the speed of the plane and the tension of the string if the angle to the vertical is 30˚.

FG

FT

60º

0.44cos60r = 0.22 m

L = 0.44 mm = 0.075 kg

UCM

= 1.1 m/s

FG

FT

60º

2cosx T

mvF F

r

siny TF F mg

2

sin

cosT

T

F mgr

F mv

2

tangr

v

tan

grv

(9.8)(0.22)

tan 60

L = 0.44 mm = 0.075 kgr = 0.22 m

= 0.85 N sinTmg

F

0.075(9.8)

sin 60

siny TF F mg

4.15 In a Rotor ride at the amusement park the room radius is 4.6 m and the rotation frequency is 0.5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

mg NF

frF

UCM

r = 4.6 mf = 0.5 rev/sT = 2 s

r = 4.6 mf = 0.5 rev/sT = 2 s

mg NF

frF

fr 0 yF F mg frF mg

fr N sF F mg Ns

mgF

2

N

mvF

r

2 rv

T

2

2

4N

mrF

T

2

2

4 N

s

mr mgF

T

2

2

4s

gT

r

2

2

9.8 2

4 4.6 = 0.22

MOTION IN A VERTICAL CIRCLE

When a body moves in a vertical circle at the end of a string, the tension FT in the string

varies with the body's position. The centripetal force Fc on the body at any point is the vector sum of FT

and the component of the body's weight toward the center of the circle.

At the highest point in the circular turn the centripetal force is given by:

At the lowest point in the circular turn the centripetal force is given by:

2

2T

mvF mg

R

2

1T

mvF mg

R

4.16 A string 0.5 m long is used to whirl a 1-kg stone in a vertical circle at a uniform velocity of 5 m/s. a. What is the tension in the string when the stone is at the top of the circle

r = 0.5 mm = 1 kgv = 5 m/s

Top of CircleFc = FT + Fg

FT = Fc - Fg

2mvmg

r

2(1)(5)1(9.8)

0.5

= 50 - 9.8 = 40.2 N

UCM

b. What is the tension in the string when the stone is at the bottom of the circle?

Bottom of CircleFc = FT - Fg

FT = Fc + Fg

= 50 + 9.8 = 59.8 N

Example 12: What is the apparent weight of a 60-kg person as she moves through the highest point when R = 45 m and the speed at that point

is 6 m/s?

n

mg

+

R

v

v

mg - n = mv 2

Rn = mg -

mv 2

R

Apparent weight will be the normal force at the

top:

22 (60kg)(6m/s)

60 kg(9.8 m/s )45 m

n n = 540 Nn = 540 N

For Motion in Circle

R

v

v

AT TOP:

T

mg

+

AT BOTTOM:

Tmg

+

T = - mg mv2

R

T = + mg

mv2

R

The force on you is the leastThe force on you is the least

The force on you is the greatest The force on you is the greatest

Summary: Ferris Wheel

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

mg - n = mv 2

R

n = + mg

mv 2

R

R

v

vn = mg -

mv 2

R

The Loop-the-Loop

AT TOP:

FN

mg+

AT BOTTOM:

FN

mg+

FN = - mg mv2

R

FN = + mg mv2

R

R

v

v

The Ferris Wheel

AT TOP:FN

mg

+

AT BOTTOM:FN

mg+

mg - FN = mv2

R

FN = + mg mv2

R

R

v

vFN = mg -

mv2

R

4.17 A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical circle, as shown. When the ball is at point P, the string is horizontal. Point Q is at the bottom of the circle and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the given quantities and fundamental constants.

UCM

a. On the figures below, draw and label all the forces exerted on the ball when it is at points P and Q.

FT

FG

FT

FG

FG

The minimum speed will occur when there is no FT

2mvmg

r

b. Derive an expression for vmin the

minimum speed the ball can have at point Z without leaving the circular path.

2

G

mvF

r

v Rg

c. The maximum tension the string can have without breaking is Tmax Derive an expression for

vmax, the maximum speed the ball can have at

point Q without breaking the string.

2

T G

mvF F

r

2

max

MvT Mg

R

2max( )R T Mg Mv

max( )R T Mgv

M

d. Suppose that the string breaks at the instant the ball is at point P. Describe the motion of the ball immediately after the string breaks.

The ball will go straight up i.e. tangential to point P

ExamplesThe maximum tension that a 0.50

m string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle?

mgT

smvm

mgTrv

mvmgTrr

mvmgT

r

mvmaFF ccNET

/74.525.0

))8.9)(25.0(14(5.0)(

)( 22

2

Examples

At the bottom?

smvm

mgTrv

mvmgTrr

mvmgT

r

mvmaFF ccNET

/81.425.0

))8.9)(25.0(14(5.0)(

)( 22

2

mg

T

Like a ball on a string, Satellites are an example of uniform circular motion.

There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

The gravitational pull of the Earth provides the centripetal force and acts like an invisible guideline for a satellite.

Fc = mac

r

GMv E

Example 4: A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine:a. The height above the Earth’s surface such a satellite must orbit and

mE = 5.98x1024 kg

RE = 6.38x106 m

T = 1 day = 86400 s

FUG = FC

r

vm

r

MGm S

E

ES2

2 T

rv

2

2

2

2

22

2

44

T

r

rT

r

r

GM E GMET2 = 4 π2r3

2

23

4TGM

r E2

22411

4

)86400)(1098.5(1067.6

skgxx

r = 4.23x107 m from the Earth’s center

b. The satellite’s speed

height = r - rE

= 4.23x107 - 6.38x106 m= 3.592x107 m

r

vm

r

MGm S

E

ES2

2

r

GMv E

mx

kgxx7

2411

1023.4

)1098.5(1067.6

= 3070 m/s

ExamplesVenus rotates slowly about its

axis, the period being 243 days. The mass of Venus is 4.87 x 1024 kg. Determine the radius for a synchronous satellite in orbit around Venus. (assume circular orbit)

Fg

3

2

272411

32

2

2

23

2

22

2

2

2

4

)101.2)(1087.4)(1067.6(

44

4

2

xxxr

GMTr

GMTr

T

r

r

GM

T

rvv

r

GMr

mv

r

MmGFF

c

cg

1.54x109 m

Kepler’s Laws and Newton's Synthesis

Kepler’s laws describe planetary motion.

1. The orbit of each planet is an ellipse, with the Sun at one focus.

2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

Figure 6.30

• The shaded regions have equal areas. It takes equal times for m to go from A to B, from C to D, and from E to F. The mass m moves fastest when it is closest to M . Kepler’s second law was originally devised for planets orbiting the Sun, but it has broader validity.

As the planet is closest the sun, the planet is moving fastest and as the planet is farthest from the sun,it is moving slowest. Nonetheless, the imaginary line adjoining the center of the planet to the center of the sun sweeps out the same amount of area in each equal interval of time.

KEPLER’S THIRD LAW"If T is the period and r is the length of the semi-major axis of a planet’s orbit, then the ratio T2/r3 is the same for all planets."

T

T

r

r12

22

13

23

The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun: T2/r3

Kepler’s laws can be derived from Newton’s laws. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our Solar System.

Newton’s Law of Universal Gravitation

If the force of gravity is being exerted on objects on Earth, what is the origin of that force?

Newton’s realization was that the force must come from the Earth.

He further realized that this force must be what keeps the Moon in its orbit.

Newton’s Law of Universal Gravitation

The gravitational force must be proportional to both masses. By observing planetary orbits, Newton concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.

In its final form, the Law of Universal Gravitation reads:

Where:

The magnitude of the gravitational constant G can

be measured .

This is the Cavendish experiment.

4.18 The mean distance from the Earth to the Sun is 1.496x108 km and the period of its motion about the Sun is one year. The period of Jupiter’s motion around the Sun is 11.86 years. Determine the mean distance from the Sun to Jupiter.

rE = 1.496x108 km

TE = 1 year

TJ = 11.86 years

3

3

2

2

J

E

J

E

r

r

T

T

3/1

2

32

E

EJJ T

rTr

3/1

2

382

)1(

)10496.1()86.11(

x= 7.77x108 km

KL

4.19 Derive Kepler’s Third Law from Newton’s Law of Gravitation. For a Planet 1 of mass m1 and the Sun of mass MS

ΣF = ma = mac

1

211

21

1

r

vm

r

MGm S 1

11

2

T

rv

substituting:2

1

12

21

4

T

r

r

GM S

rearranging:

SGMr

T 2

31

21 4

For a Planet 2 of mass m2 and the Sun of

mass MS: SGMr

T 2

32

22 4

therefore:3

2

1

2

2

1

r

r

T

T

ULG/KL

4.20 What is the force of gravity acting on a 2000 kg spacecraft when it orbits two Earth radii from the Earth’s center above the Earth’s surface?

m1 = 2000 kg

ME = 5.98x1024 kg

rE = 6380x103 (2) =1.276x107 m

221

r

mGmF

27

2411

)10276.1(

)1098.5)(2000(1067.6

x

xx

= 4899 N

ULG

4.21 a. Derive the expression for g from the Law of Universal Gravitation.

Fg = FUG

2E

E

r

GmMmg

2E

E

r

GMg

ULG

b. Estimate the value of g on top of the Everest (8848 m) above the Earth’s surface.

mE = 5.98x1024 kg

RE = 6.38x106 m

RT = 8848 + 6.38x106 = 6.388x106 m

2E

E

r

GMg 26

2411

)10388.6(

)1098.5(1067.6

x

xx

= 9.77 m/s2

The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

Satellites

Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.

The satellite is kept in orbit by its speed – it is continually free falling, but the Earth curves from underneath it.

4.22 A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine:a. The height above the Earth’s surface such a satellite must orbit

mE = 5.98x1024 kg

RE = 6.38x106 m

T = 1 day = 86400 s

FUG = FC

r

vm

r

MGm S

E

ES2

2 T

rv

2

2

2

2

22

2

44

T

r

rT

r

r

GM E GMET

2 = 4 π2r3

2

23

4TGM

r E2

22411

4

)86400)(1098.5(1067.6

xx

r = 4.23x107 m from the Earth’s center

UCM

b. The satellite’s speed

height = r - rE

= 4.23x107 - 6.38x106 m = 3.592x107 m

r

vm

r

MGm S

E

ES2

2

r

GMv E

7

2411

1023.4

)1098.5(1067.6

x

xx

= 3070 m/s

Figure 6.24

• The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the far side. The distances and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per day at any location, because Earth rotates under the tidal bulge.

Figure 6.25

• (a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the Sun lies at 90º to the Earth-Moon alignment. Note that this figure is not drawn to scale.

Figure 6.26

• A black hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one star in a two-star system. The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth.

Figure 6.27

• Astronauts experiencing weightlessness on board the International Space Station. (credit: NASA)