On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 {\rm km} above the earth's surface, and we shall assume a circular orbit.
How many hours does it take this satellite to make one orbit? (T=?hours)
How fast (in {\rm km/s} )is the Aura spacecraft moving? (v=?km/s)
3 answers
A boat propelled so as to travel with a speed of 0.50m/s in still water, moves directly ( in a straight line) across the river that is 60m wide. The rivers flows with a speed of 0.30m/s. How long in a seconds does it take the boat to across the river?
On the earth surface g = GM/R^2
And at the height h
g" = GM/ [R+h]^2
g" = R^2/ [ R+h]^2•g
g" = {R/ [ R+h] }^2•g
g" = {6378/ 7083}^2•9.81 = 7.95 m/s^2
T^2 = 4π^2•(R+h) /g"
T^2 = 4π^2•7083 /7.95
T = 187.54s = 0.052 hour.
v = 2•π •[R+h] / T
v = 2•π •[7.083] / 187.54s
v = 0.24 km /s
And at the height h
g" = GM/ [R+h]^2
g" = R^2/ [ R+h]^2•g
g" = {R/ [ R+h] }^2•g
g" = {6378/ 7083}^2•9.81 = 7.95 m/s^2
T^2 = 4π^2•(R+h) /g"
T^2 = 4π^2•7083 /7.95
T = 187.54s = 0.052 hour.
v = 2•π •[R+h] / T
v = 2•π •[7.083] / 187.54s
v = 0.24 km /s
I've mistaken in the previous post.
New version:
m•a(norm) = GM/ [R+h] ^2,
m•v^2/(R+h) = GM/ [R+h] ^2,,
v = sqrt[GM/(R+h)] =sqrt [6.67•10^-11•5.97•10^24/7.1•10^6] =
= 7490 m/s =7.49 km/s
T = 2 •π(•R+h)/v =2 •π•7.1•10^6/7490= =5960 s =1.65 h
New version:
m•a(norm) = GM/ [R+h] ^2,
m•v^2/(R+h) = GM/ [R+h] ^2,,
v = sqrt[GM/(R+h)] =sqrt [6.67•10^-11•5.97•10^24/7.1•10^6] =
= 7490 m/s =7.49 km/s
T = 2 •π(•R+h)/v =2 •π•7.1•10^6/7490= =5960 s =1.65 h
1 answer