Question

satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is closest to the Earth, it is a distance of1.00×106m from the surface (not the center) of the Earth, andis moving at a velocity of5.14km/s. At the point in its orbit when it is furthest from the Earth it is a distance of2.00×106m from the surface of the Earth. Note that the Earth has a mass of5.97×1024kgand a radius of6.37×106m.(a)How fast is the satellite moving when it is at its furthest point from the Earth?(b)If, at the closest approach to the Earth, the satellite could be deflected so that it had the same velocity as before,but was now traveling directly away from the Earth, how far from the surface of the Earth would it get before itstopped, and then started to fall back to the Earth?

Answer 1

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Step-by-step explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m