An artificial earth satellite has a circular orbit of radius 5.91 X 10^6 m (whic

Question

An artificial earth satellite has a circular orbit of radius 5.91 X 10^6 m (which means it is orbiting approximately -450 km above the surface of the earth) in an equatorial plane. The period T (the time required for one complete orbit) is 5.31 X 10^3 s (about 1.48 h). (a) Compute the (constant) speed of the satellite. (b) If the satellite is directly above the equator and traveling east at time t, find the average acceleration during the time interval from t to t + T/40.0. Magnitude? Direction?(c) Find the satellite's instantaneous acceleration at time t. Magnitude? Direction?

Explanation / Answer

a)Speed of satellite= 2r/T=2*5.91*106/(5.31*103)=6993m/s =6.99 km/s

b)The angle covered by satellite in T/40=2/40=/20 radians.

This is also the change in direction of the satellite.

Change in velocity= 6.99cos(/20) -6.99 tangentially and 6.99*sin(/20) radially.

=-0.086 tangentially ,1.09 radially.

Average accleration=(-0.086i+1.09j)/(5310/40)=-6.48*10-4km/s2 tangentially and 8.21*10-3km/s2 radially.

magnitude=8.24*10-3 km/s2

c) instantaneous acceleation= v2/r radially= 6.992/(5.91)=8.27 km/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.