An Earth satellite moves in a circular orbit 685 km above the Earth's surface. The period of the motion is 98.3 min.
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?
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Average speed is the distance traveled divided by the time required to cover the distance. It is a scalar quantity since it has no direction. V= aver displacement/aver. time
Average acceleration is the change in velocity from an initial value v0 to the final velocity v divided by the elapsed time. a= aver. V/aver. t
I think in this problem you are trying to obtain the average velocity which is displacement divided by the time, in physics the metric system uses seconds so I recommend you change the minutes to seconds there.
Also I think that you are dealing with uniform circular motion which in order to obtain the velocity of this example you need to find the radius between the earth and the satellite. The thing is that there is only one speed that the satellite can have if the satellite is to remain in orbit with a fixed radius.
The equation would be"
v= √ GME /r
G is the universal gravitational constant: 6.67X10^-11 N m^2/kg^2
ME it is suppose to be a smaller E but couldn't do it on my computer is the mass of the earth
r is the distance from the center of the earth to the satellite. Th radius of the earth is approx. 6.38X10^6m
The equation for centripetal acceleration is ac= v^2/r, so once you have gotten the radius between the satellite and the earth, you can obtain this with the height where the satellite is located, you will be able to substitute the values and get your answer.
Average acceleration is the change in velocity from an initial value v0 to the final velocity v divided by the elapsed time. a= aver. V/aver. t
I think in this problem you are trying to obtain the average velocity which is displacement divided by the time, in physics the metric system uses seconds so I recommend you change the minutes to seconds there.
Also I think that you are dealing with uniform circular motion which in order to obtain the velocity of this example you need to find the radius between the earth and the satellite. The thing is that there is only one speed that the satellite can have if the satellite is to remain in orbit with a fixed radius.
The equation would be"
v= √ GME /r
G is the universal gravitational constant: 6.67X10^-11 N m^2/kg^2
ME it is suppose to be a smaller E but couldn't do it on my computer is the mass of the earth
r is the distance from the center of the earth to the satellite. Th radius of the earth is approx. 6.38X10^6m
The equation for centripetal acceleration is ac= v^2/r, so once you have gotten the radius between the satellite and the earth, you can obtain this with the height where the satellite is located, you will be able to substitute the values and get your answer.
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(a) After each revolution, the satellite sweeps out a distance of:
C = 2πr
= 2π(685 + 6368 km)
= 14106π km.
Since the satellite does this every 98.3 min = 5898 s, the speed of the satellite is:
v = d/t
= (14106π km)/(5898 s)
= 7.51 km/s or 7.51 x 10^3 m/s, to 3 s.f.
(b) Since the planet travels in a circle, its centripetal acceleration is:
a(c) = v^2/r
= (7.51 x 10^3 m/s)^2/(6.85 x 10^5 m + 6.37 x 10^6 m)
= 7.99 m/s^2, to 3 s.f.
I hope this helps!
C = 2πr
= 2π(685 + 6368 km)
= 14106π km.
Since the satellite does this every 98.3 min = 5898 s, the speed of the satellite is:
v = d/t
= (14106π km)/(5898 s)
= 7.51 km/s or 7.51 x 10^3 m/s, to 3 s.f.
(b) Since the planet travels in a circle, its centripetal acceleration is:
a(c) = v^2/r
= (7.51 x 10^3 m/s)^2/(6.85 x 10^5 m + 6.37 x 10^6 m)
= 7.99 m/s^2, to 3 s.f.
I hope this helps!
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Speed of satellite
V² = GM(of earth)/R (of earth) + h
V² = 6.67 x 10^-11 x 6 x 10^24 / 6.4 x 10^6 + 6.85 x 10^5
V² = 5650000
V = 7516m/s
Centripetal acceleration
A = V²/R(of earth) + h
A = 8m/s²
V² = GM(of earth)/R (of earth) + h
V² = 6.67 x 10^-11 x 6 x 10^24 / 6.4 x 10^6 + 6.85 x 10^5
V² = 5650000
V = 7516m/s
Centripetal acceleration
A = V²/R(of earth) + h
A = 8m/s²