Download AST 341 Final Hazırlık Soruları 1. (a) The asteroid Pallas has an

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AST 341 Final Hazırlık Soruları
1. (a) The asteroid Pallas has an average diameter of 520 km and a mass of 3.2 1020 kg.
How much would a 100-kg astronaut weigh there? (b) What is the asteroid’s escape
speed?
1. (a) We can calculate the acceleration due to the surface gravity using the equation
a
GM p
Rp2
2
(6.67 1011 )(3.2 1020 )
m/s
.


0.32
(260, 000) 2
This is 31 times smaller than the Earth’s surface gravity, so a 100 kg person would weigh
3.2 kg on Pallas.
(b) The formula for escape speed is
v
2GM
= 405 m/s = 0.41 km/s
r
2. You are standing on the surface of a spherical asteroid 10 km in diameter, of density
3000 kg/m3. Could you throw a small rock fast enough that it escapes? Give the speed
required in km/s
2. To calculate the escape velocity, we must know the mass, which we can calculate from
4
the density and volume. Mass = density × volume = 3000   (5000)3  1.6 1015 kg.
3
Now to the equation:
v
2GM
= 6.47 m/s = .0065 km/s
r
3. (a) NEAR’s initial orbit around Icarus had periapsis (distance of closest approach) of
about 100 km from the asteroid’s center. If the mass of Eros is 6.7 1015 kg and the orbit
had an eccentricity of 0.3, calculate the spacecraft’s orbital period. (b) Subsequently NEAR
moved into a closer orbit, with periapsis 14 km and apoapsis (greatest distance) 60 km.
What was the new orbital period?
3. We can find these orbital periods from Kepler’s Third Law: P 2 
a3
. The mass of NEAR
M Tot
is small, so M is purely the mass of Icarus in solar masses, 3.37 1015 solar masses.
a. We can find the satellite’s semi-major axis from the perihelion distance and
the eccentricity:
100 = a (1 - 0.3), for a = 143 km = 0.00000095 AU.
(0.00000095)3
So P 2 
and P = 0.016 years or 5.8 days.
3.37 1015
b. The semi-major axis is the average of the two extremes, or 37 km, which is
0.00000025 AU.
(0.00000025)3
and P = 0.0021 years or 18.5 hours.
P2 
3.37 1015
4. Stellar radial velocity variations as small as 1 m/s can be detected with current
technology. For a jupiter-mass planet orbiting a Sun-like star, this corresponds to a
planetary orbital velocity of approximately 1 km/s. Based on this information, what is the
radius of the widest circular orbit on which jupiter could currently be detected orbiting the
Sun?
4. We can find the distance from the formula
11
30
GM
GM  6.67 10 1.99 10 
v
so r  2 
 1.33 1014 m
2
r
v
1000 
This is almost 900 AU! We should be able to detect many more “Jupiters” in the future.