The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An obje
ct is launched straight up from just above the atmosphere of Venus. (a) What initial speed is needed so that when the object is far from Venus its final speed is 8000 m/s? vinitial = m/s (b) What initial speed is needed so that when the object is far from Venus its final speed is 0 m/s? (This is called the "escape speed.") vescape = m/s
1 answer:
Answer:
(a) The initial speed required is 13116 m/s
(b) The escape speed is 10394 m/s
This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.
The full calculation can be found in the attachment below.
Explanation:
In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.
The calculation can be found below.
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<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>
Explanation:
a) Acceleration due to gravity
Here we need to find acceleration due to gravity of Sun,
G = 6.67259 x 10⁻¹¹ N m²/kg²
Mass of sun, M = 1.989 × 10³⁰ kg
Radius of sun, r = 6.957 x 10⁸ m
Substituting,
Acceleration due to gravity on the surface of the Sun = 274.21 m/s²
b) Acceleration due to gravity in earth = 9.81 m/s²
Ratio of gravity = 274.21/9.81 = 27.95
Weight = mg
Factor of increase in weight = 27.95
Answer:
Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let
denote the speed of light in the air. Assume that the light entered the boundary at an angle of
to the normal and exited with an angle of
. By Snell's Law, the sine of
and
would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now
. (in this case,
.)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For ,
is monotonically increasing with respect to
. In other words, for
in that range, the value of
increases as the value of
increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle
would have a larger
. such that light would travel faster in that medium.