Answer:
a) v = √ G M / r, , b) K = ½ m G M / r, , c) T = 2 π √ (r³ / GM)
, d) L = m √ (rGM)
Explanation:
The formula of universal gravitation
F = G m M / r²
Part A) orbital speed
Let's use Newton's second law
F = m a
Where the acceleration is centripetal
a = v² / r
Let's replace
G m M / r² = m v² / r
v = √ G M / r
Part B)
Kinetic energy
K = ½ m v²
K = ½ m G M / r
Part C)
The orbital period The speed (speed module) is constant, so we can use the relationships
v = d / t
The distance in the length of the circle
d = 2π r
T = t = 2π r / (√ G M / r) = 2π r √(r / GM)
T = 2 π √ (r³ / GM)
Part D) The angular momentum
L = I w
The satellite is small, so we can approximate it to a particle
I = m r²
The angular and linear velocity are related
v = w r
w = v / r
w = √(G M / r) (1 / r)
We replace
L = m r² 1 / r √(GM / r)
L = m r √ (GM / r)
L = m √ (rGM)
Part E) let's calculate the order of magnitude of the quantities
v = √G M / r
The value of the distance from the plant center to the satellite
r = Re + h
the height of the satellites is less than 1 10⁶m from the surface of the earth,
r = 6.37 10⁶ + 1 10⁶ = 7.37 10⁶
v = √ (6.67 10⁻¹¹ 5.98 10²⁴ / 7.37 10⁶)
v = √ (5.4 10⁶)
v ~ 10³ m / s
K = ½ m G M / r
The satellite mass about 10⁴ kg
K = ½ 10⁴ 6.67 10⁻¹¹ 5.98 10²⁴ / 7.37 10⁶
K ~ 10¹¹ J
U = G m M / r
U ~ 10¹¹ J
L = m √ (rGM)
L = 10⁴ √ (7.37 10⁶ 6.67 10⁻¹¹ 5.98 10²⁴)
L ~ 10¹⁴ kg m2 / s
