You are the pilot of a spaceship (9.1835 × 104 kg) and you enter into elliptical orbit around an uncharted planet, Rosseforp. At
periapsis (closest approach to the planet), your ship is 4.5 × 107 m above the surface of the planet; at apoapsis (most distant) it is 2.257×108 m above the surface. The mass of Rosseforp is found by your ship’s equipment to be 1.96×1025 kg, and the period of your ship’s orbit is 83.37 hours
Find the following:i. the semimajor axis of your ship’s orbit,ii. the radius of planet Rosseforp,iii. the eccentricity of your ship’s orbit,iv. the speed of your ship at perihelion and aphelion,
Find the following:i. the semimajor axis of your ship’s orbit,ii. the radius of planet Rosseforp,iii. the eccentricity of your ship’s orbit,iv. the speed of your ship at perihelion and aphelion,
1 answer:
Answer:
Check the explanation
Explanation:
The orbital period of a satellite that is given as (T) and the mean distance from the central body (R) are connected by the following equation: representing T as the satellite period, R will be represented as the average radius of orbit for the satellite (which is the distance from center of central planet)
Kindly check the attached image below to see the step by step explanation to the above question.
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Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
Complete Question:
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.
Answer:
a) Speed of car A at the start of sliding = 4.23 m/s
b) speed of car B at the start of sliding = 3.957 m/s
c) Speed of car B before the collision = 7.28 m/s
Explanation:
NB: The figure is not provided but all the parameters needed to solve the question have been given.
Let the frictional force acting on car A, ............(1)
Since frictional force is a type of force, we are safe to say .......(2)
Equating (1) and (2)
a) Speed of A at the start of the sliding
b) Speed of B at the start of the sliding
Let the speed of car B before collision =
Momentum of car B before collision =
Momentum after collision =
Applying the law of conservation of momentum: