Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's surface, while that for satellite b is at a height of 838 km. find the orbital speed for satellite a and satellite b.

Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite 
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g' 
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h): 
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)] 
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec 
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec

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Velocity=total displacement/timefrom house to store is 500 from store to neighbor is 200 therefore total displacement=500+200=700metrestherefore,700metres/910seconds=0.8metres per second