Rotary Homework #4

Rotary Homework #4: 1. At the surface of a certain planet, the gravitational acceleration g has a magnitude of 12.0 m s 2 . A 21.0-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?

2. Calculate the force of Earth’s gravity on a spacecraft 12,800 km (2 Earth radii) above 24 the Earth’s surface if its mass is 1350 kg. RE = 6378 km and ME = 5.98 x 10 kg

3. Calculate the acceleration due to gravity on the Moon. The Moon’s radius is 1.74106 m and its mass is 7.3510 22 kg.

4. A hypothetical planet has a radius 1.5 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?

5. A hypothetical planet has a mass 1.66 times that of Earth, but the same radius. What is g near its surface?

6. Two objects attract each other gravitationally with a force of 2.510 10 N when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual masses.

7. A certain neutron star has five times the mass of our Sun packed into a sphere about 30 10 km in radius. Estimate the surface gravity on this monster. MSun = 1.99 x 10 kg

8. Objects with masses of 200 kg and 500 kg are separated by 0.400 m. (a) Find the net gravitational force exerted by these objects on a 50.0-kg object placed midway between them. (b) At what position (other than infinitely remote ones) can the 50.0-kg object be placed so as to experience a net force of zero? Rotary Homework #5: 9. A satellite moves in a circular orbit around Earth at a speed of 5,000 m/s. Determine (a) the satellite’s altitude above the surface of Earth and (b) the period of the satellite’s 24 orbit. RE = 6378 km and ME = 5.98 x 10 kg

10. A 600-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius. Find (a) the satellite’s orbital speed, (b) the period of its revolution, and (c) the gravitational force acting on it.

11. Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 × 105 km. From these data, determine the mass of Jupiter.

12. Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose that the mass of a certain spherical neutron star is twice the mass of the sun and its radius is 10.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its 30 surface on the equator is just held in orbit by the gravitational force. MS = 1.99 x 10 kg

13. Assume that you are agile enough to run across a horizontal surface at 8.50 m/s, independently of the value of the gravitational field. What would be (a) the radius and (b) the mass of an airless spherical asteroid of uniform density 1.10 × 103 kg/m3 on which you could launch yourself into orbit by running? (c) What would be your period? {HINT: Write Kepler’s 3rd Law in terms of the density of the gravitating source rather than the mass.} m 4 mv 2 GMm density = D  , volume of a sphere = V   R 3 , centripetal force =  V 3 R R 2 Solutions: 1. (a) Mass is independent of location and so the mass of the ball is 21.0 kg on both the Earth and the planet. (b) The weight is found by W= mg . 2 WEarth= mg Earth =(21.0 kg)( 9.80 m s) = 206 N

2 WPlanet= mg Planet =(21.0 kg)( 12.0 m s) = 252 N .

2. The spacecraft is three times as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force as 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. 1350 kg 9.80 m s 2 1 ( )( ) 3 FG =9 mgEarth's = = 1.47 10 N surface 9

3. The force of gravity on an object at the surface of a planet is given by Newton’s law of Universal Gravitation, using the mass and radius of the planet. If that is the only force on an object, then the acceleration of a freely-falling object is acceleration due to gravity.

MMoon m FG = G2 = mgMoon rMoon 22 M (7.35 10 kg) Moon 醋 -11 2 2 2 gMoon = G =(6.67 10 N m kg) 2 = 1.62 m s r 2 6 Moon (1.74 10 m)

4. The acceleration due to gravity at any location on or above the surface of a planet is given by 2 gplanet= G M Planet r , where r is the distance from the center of the planet to the location in question. M M1 M 1 9.8 m s2 g= GPlanet = G Earth = G Earth = g = = 4.4 m s2 planetr22 1.5 2 R 2 1.5 2 Earth 1.5 2 (1.5REarth ) Earth

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

MPlanet1.66 M Earth骣 M Earth 2 2 gplanet= G2 = G 2 =1.66琪 G 2 = 1.66 g Earth = 1.66( 9.80 m s) = 16.3m s r REarth桫 R Earth

6. Assume that the two objects can be treated as point masses, with m1  m and m2 4 kg  m . The gravitational force between the two masses is given by m m m(4 - m) 4m- m2 1 2 醋 -11 2 2 - 10 F= G2 = G 2 =(6.67 10 N m kg) 2 = 2.5 10 N . r r (0.25 m) This can be rearranged into a quadratic form of m2 -4 m + 0.234 = 0 . Use the quadratic formula to solve for m, resulting in two values which are the two masses.

m1=3.9 kg , m 2 = 0.1 kg .

7. The acceleration due to gravity at any location at or above the surface of a star is given by 2 gstar= G M star r , where r is the distance from the center of the star to the location in question. 5 1.99 1030 kg Mstar5 M Sun -11 2 2( ) 12 2 gstar = G2 = G 2 = (6.67� 10 Ng m kg) 2 7 10 m s r r (1 104 m)

8. (a) At the midpoint between the two masses, the forces exerted by the 200-kg and 500-kg Gm m masses are oppositely directed, and from F = 1 2 we have r2

G(50.0 kg)( 500 kg- 200 kg ) SF = = 2.50 10-5 N toward the 500-kg (0.200 m )2

(b) At a point between the two masses and distance d from the 500-kg mass, the net force will be zero when

G(50.0 kg)( 200 kg) G( 50.0 kg)( 500 kg ) = or d = 0.245 m (0.400 m - d)2 d2

Note that the above equation yields a second solution d = 1.09 m . At that point, the two gravitational forces do have equal magnitudes, but are in the same direction and cannot add to zero.

9. (a) The gravitational force must supply the required centripetal acceleration, so

Gm m 骣v 2 Gm E t r = E 2 = m 琪 . This reduces to 2 , which gives r桫 r vt

2 5.98 1024 kg 骣 -11N m ( ) 7 r = 琪6.67� 10 2 2 1.595 10 m 桫 kg (5 000 m s)

The altitude above the surface of the Earth is then

7 6 6 h= r - RE = 1.595� 10 m � 6.38 10 m 9.57 10 m (b) The time required to complete one orbit is

7 circumference of orbit 2p ( 1.595 10 m ) T = = = 2.00� 104 s 5.57 h orbital speed 5 000 m s

10. (a) The satellite moves in an orbit of radius r= 2 RE and the gravitational force supplies 2 2 the required centripetal acceleration. Hence, m( vt2 R E) = Gm E m( 2 R E ) , or

24 Gm 骣 N m 2 (5.98 10 kg ) v =E = 6.67 10-11 5.59 10 3 m s t 琪 � 2 6 2RE 桫 kg 2( 6.38 10 m )

(b) The period of the satellite’s motion is

2p 轾 2 6.38 106 m 2p r 臌( ) 4 T = =3 = 1.43� 10 s 3.98 h vt 5.59 10 m s

2 (c) The gravitational force acting on the satellite is F= GmE m r , or

2 5.98 1024 kg 600 kg 骣 -11N m ( )( ) 3 F = 琪6.67� 10 2 2 1.47 10 N 桫 kg 轾 6 臌2( 6.38 10 m )

11. The gravitational force exerted on Io by Jupiter provides the centripetal 骣v 2 G M m rv 2 acceleration, so m t = , or M = t 桫琪r r2 G

The orbital speed of Io is

2p r 2p ( 4.22 108 m ) v = = = 1.73 104 m s t T (1.77 days)( 86 400 s day )

2 (4.22创 108 m)( 1.73 10 4 m s) Thus, M = = 1.90 1027 kg 6.67醋 10-11 N m 2 kg 2

12. The gravitational force on a small parcel of material at the star’s equator supplies 2 G Ms m 骣vt 2 the centripetal acceleration, or 2 =m琪 = m( Rs w ) Rs桫 R s

3 Hence, w = G Ms R s 6.67醋 10-11 N m 2 kg 2轾 2 1.99 10 30 kg ( ) 臌( ) 4 =3 = 1.63 10 rad s (10.0 103 m )

13. (a) In order to launch yourself into orbit by running, your running speed must be such that the gravitational force acting on you exactly equals the force needed to produce 2 2 the centripetal acceleration. That is, GMm r= m vt r , where M is the mass of the

骣4 3 asteroid and r is its radius. Since M= density� volumer琪 p r , this 桫3

2 2 骣4 3 m m vt 3v requirement becomes Gr琪 p r = or r = t . 桫3 r2 r 4pG r

The radius of the asteroid would then be

2 3( 8.50 m s) r = = 1.53 104 m 4p ( 6.673醋 10-11 N m 2 kg 2)( 1.10 10 3 kg m 3 )

or r = 15.3 km

(b) The mass of the asteroid is given by

骣43 3 3 4 43 16 M=r琪 p r =(1.10创 10 kg m) p ( 1.53 10 m) = 1.66 10 kg 桫3 3

(c) Your period will be

4 2p 2 pr 2p (1.53 10 m ) T = = = = 1.13 104 s w vt 8.50 m s