At a certain point in the reentry of the space shuttle into the earth's atmosphere, the total acceleration of the shuttle may be represented by two components. One component is the gravitational acceleration g = 9.75 m/s² at this altitude. The second component equals 12.80 m/s² due to atmospheric resistance and is directed opposite to the velocity. The shuttle is at an altitude of 49.3 km and has reduced its orbital velocity of 28300 km/h to 15000 km/h in the direction = 1.70°. For this instant, calculate the radius of curvature of the path and the rate at which the speed is changing. Answers: p= i = Mi km m/s²

A spy satellite is in circular orbit around Earth. It makes one revolution in 8.90 h. Mass of Earth is 5.974 × 1024 kg, radius of Earth is 6371 km and Gravitational constant G is = 6.674 × 10−11 N·m2/kg2. a. How high above Earth’s surface is the satellite? answer in km b. What is the satellite’s acceleration? answer in m/s^2

Chapter 3- questn-15 The magnitude of the centripetal acceleration of the plane moving with constant speed in a circular orbit is 9 times the gravitational acceleration. Since its constant speed is 300 m/s, find the diameter of the circular orbit. (g=10m/s²) Option 1 Option 2 Option 3 Option 4 Option 5 500 m 1000 m 2000 m 2500 m 3000 m

A projectile is shot straight up from the earth's surface at a speed of 1.00×104 km/hr .How high does it go?

A 100.0 kg rocket is launched at 8.00 km/s directly up from the earth surface. What altitude does it reach

In the figure, three 7.82 kg spheres are located at distances d₁ = 0.226 m, and d₂ = 0.204 m. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the net gravitational force on sphere B due to spheres A and C? (a) Number (b) Number IN M. d₁ A B ∙d₂ Units Units N ° (degrees)

2/113 A space shuttle which moves in a circular orbit around the earth at a height h 150 mi above its surface must have a speed of 17,369 mi/hr. Calculate the gravitational accelerationg for this altitude. The mean radius of the earth is 3959 mi. (Check your an- swer by computing g from the gravitational law g = R go where go = 32.22 ft/sec2 from Table D/2 R+h in Appendix D.)

The astroid belt of our solar system is located between the orbits of Mars and Jupiter. An astronomer is studying an interesting asteroid and has determined that one orbit around the Sun takes it 2.83 years. How far is the asteroid from the Sun? The following constants may be helpful. Earth's distance from the Sun = 1 AU = 1.5 x 1011 m MSun = 2.0 x 1030 kg G = 6.67 x 10-11 N m2 / kg2 1 year = 3.15 x 105 s A)1.5 AU (2.25 x 1011 m) B) 2.0 AU (3.00 x 1011 m) C) 2.5 AU (3.75 x 1011 m) D) 3.2 AU (4.80 x 1011 m) E) 4.8 AU (7.20 x 1011 m)

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 18.0 ° north of the equator. Re (a) v= i Units ac = Units (b) v = i Units ac = Units > > >

An Earth satellite moves in a circular orbit 687 km above Earth's surface with a period of 98.30 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?