More Examples About Gravity Stuff Answers 1. Find the value of g at an altitude of m above Earth’s surface. ReRe r g = G M e r2r m r = R e m = ( 6.37 x 10 6 m ) m = m g = G M e r2r2 = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) ( m ) 2 g = 9.68 m/s 2

2. Find the value of g in Denver, 1760 m above sea level. ReRe r g = G M e r2r m r = R e m = ( 6.37 x 10 6 m ) m = m g = G M e r2r2 = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) ( m ) 2 g = 9.79 m/s 2

3. Find the value of g at an altitude of 350 km above Earth’s surface. 350 km ReRe r g = G M e r2r2 = m r = R e m = ( 6.37 x 10 6 m ) m = m g = G M e r2r2 = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) ( m ) 2 g = 8.80 m/s 2

4. Find the value of g at a location 3 Earth radii from Earth’s center. ReRe 3R e g = G M e r2r2 r = 3R e = 3( 6.37 x 10 6 m ) g = G M e r2r2 = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) ( x 10 7 m ) 2 g = 1.09 m/s 2 = x 10 7 m

5. A satellite of mass 3000 kg is orbiting 500 km above Earth’s surface. Determine its horizontal speed. 500 km ReRe r = R e km v = G M e r = ( 6.37 x 10 6 m ) m = m v = G M e r = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) m v = 7607 m/s

6. The ISS has an orbital altitude of 350 km. Find the horizontal speed necessary for stable orbit. 350 km ReRe r = R e + 350km v = G M e r = ( 6.37 x 10 6 m ) m = m v = G M e r v = 7691 m/s = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) m

7. Find the horizontal velocity for a satellite 3 Earth radii away from Earth’s center. ReRe 3R e v = G M e r r = 3R e = 3( 6.37 x 10 6 m ) = x 10 7 m v = G M e r v = 4561 m/s = ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg ) x 10 7 m

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) A geosynchronous satellite has an orbital period of 24 hours, the same as Earth ReRe ? r T = 24 hr r = ? v = G M e r Can’t use since we know nothing about the speed at that altitude Need to start over and use centripetal force arguments m1m1

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) A geosynchronous satellite has an orbital period of 24 hours, the same as Earth ReRe ? r T = 24 hr r = ? Centripetal force needed for circular motion is provided by gravity F c = F g This time, use for F c 4 π 2 m 1 r T2T2 m1m1

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) A geosynchronous satellite has an orbital period of 24 hours, the same as Earth ReRe ? r T = 24 hr r = ? m1m1 F c = F g 4 π 2 m 1 r T2T2 = G m 1 M e r2r2 Need to solve this for r, the distance from Earth’s center; to eliminate denominators, cross- multiply 4 π 2 m 1 r 3 = G m 1 M e T 2 4 π 2 r 3 = G M e T 2

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) A geosynchronous satellite has an orbital period of 24 hours, the same as Earth ReRe ? r T = 24 hr r = ? m1m1 4 π 2 r 3 = G M e T 2 1 hr 60 min60 s 1 min T = s r 3 = G M e T 2 4 π 2 r = 3 G M e T 2 4 π 2

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) ReRe ? r m1m1 T = s r = 3 G M e T 2 4 π 2 = 3 ( 6.67 x Nm 2 /kg 2 )( 5.96 x kg )( s ) 2 4 π 2 r = m = x m 3 r = 4.22 x 10 7 m This is distance from Earth’s center; to find altitude, subtract Earth’s radius

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) ReRe ? r r = 4.22 x 10 7 m Altitude = ( 4.22 x 10 7 m ) – ( 6.37 x 10 6 m ) ReRe Altitude = 3.58 x 10 7 m How many times Earth’s radius is this altitude? 3.58 x 10 7 m 6.37 x 10 6 m = 5.62

8. Determine the altitude of a geosynchronous satellite (a satellite that continually stays above the same location on Earth) ReRe 6.6 R e r Altitude = 5.6 Earth radii or, distance from Earth’s center = 6.6 Earth radii geosynchronous satellites are way out there

9. Find the period of a pendulum 1.50 m long (a) on Earth’s surface T = L g 2 π 1.50 m 9.80 m/s 2 = 2 π T = 2.46 s

9. Find the period of a pendulum 1.50 m long (b) on the Lunar surface T = L g 2 π 1.50 m 1.62 m/s 2 = 2 π T = 6.05 s g = G M m Rm2Rm2 = ( 1.74 x 10 6 m ) 2 ( 6.67 x Nm 2 /kg 2 )( 7.35 x kg ) g = 1.62 m/s 2

10. Find the length of a Foucault pendulum that has a period of 7.50 s. T = L g 2 πSolve for L 2 π L g T = ( ) 2 T2T2 4 π 2 = L g g T 2 = 4 π 2 L 4 π 2 L = g T 2 ( 9.80 m/s 2 )( 7.50 s ) 2 4 π 2 L = 4 π 2 = L = 14.0 m