Answer:
(a) g = 8.82158145.
(b) 7699.990192m/s.
(c)5484.3301s = 1.5234 hours.(extremely fast).
Explanation:
(a) Strength of gravitational field 'g' by definition is
, here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.
r = 6721,000 meters, putting this value in above equation gives g = 8.82158145.
(b) We have to essentially calculate centripetal acceleration that equals new 'g'.
here g is known, r is known and v is unknown.
plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.
(c) S = vT, here T is time period or time required to complete one full revolution.
S = earth's circumfrence , V is calculated in (B) T is unknown.
solving for unknown gives T = 5484.3301s = 1.5234hours.
Answer:
At t = 15.0 s the magnitude of the velocity is 58.31 m/s
Explanation:
The given parameters are;
V₀ₓ = 30 m/s
= 100 m/s
The time of flight of the projectile = 25 s
For projectile motion;
Vₓ = V₀ × cos(θ₀)
The magnitude of the velocity V = √(V₀ₓ² + ²)
We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s
cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109
θ₀ = cos⁻¹(3/√109) = 73.3°
The components of the velocity after time t is given by the relations;
Vₓ = V₀ × cos(θ₀) = 30 m/s
= V₀ × sin(θ₀) - g×t
When = 0, we have;
0 = V₀ × sin(θ₀) - g×t
g×t = V₀ × sin(θ₀) = 10·√109×0.958 = 100 m/s
t = 100/g = 100/10 = 10 s
The time to reach maximum height = 10 s
At 15.0 seconds, we have;
= V₀ × sin(θ₀) - g×t = 10·√109×0.958 - 10×15 = -50 m/s
Therefore, the projectile is returning at 50 m/s
The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.