If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________
1 answer:
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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Answer:
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Explanation:
The circular velocity is define as:
Where r is the radius of the trajectory and T is the orbital period
To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:
Where T is orbital period and r is the radius of the trajectory.
Case for the particle in the Encke Division:
It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( )
1 AU is defined as the distance between the earth and the sun.
But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.
That can be expressed in units of days
<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
⇒ 10247040 s
⇒ 133370000 m
Case for the particle in the D Ring:
For the case of the particle in the D Ring, the same approach used above can be followed
<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
⇒ 3786912 s
⇒ 69000000 m
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
- Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
- Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
- vₓ₀ = v * cos θ (1)
- where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
- Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:
- Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:
- ⇒θ = cos⁻¹ (0.526) = 58.3º (4)
b)
- At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
- Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
- vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)
c)
- At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)
d)
- Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:
- Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
- v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
- Replacing (7) in (6), we get:
e)
- When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
- The horizontal component, since it keeps constant, is just v₀x:
- v₀ₓ = 13.7 m/s
- The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:
- Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:
- Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows: