Respuesta :
Answer:
738.39 days.
Explanation:
To find the orbital period
T of the probe around the comet, we can use Kepler's third law of planetary motion, which relates the orbital period of a celestial body to its distance from the body it is orbiting and the mass of the body being orbited.
Kepler's third law can be expressed as:
2
=
4
2
β
β
3
T
2
=
Gβ M
4Ο
2
β
β r
3
Where:
T is the orbital period (in seconds)
G is the gravitational constant (
6.674
Γ
1
0
β
11
β
m
3
/
kg
β
s
2
6.674Γ10
β11
m
3
/kgβ s
2
)
M is the mass of the comet (
1.0
Γ
1
0
13
β
kg
1.0Γ10
13
kg)
r is the distance between the probe and the center of the comet (
30
β
km
=
30
,
000
β
m
30km=30,000m)
First, let's convert the distance from kilometers to meters:
=
30
,
000
β
m
r=30,000m
Now, we can plug the values into the equation:
2
=
4
2
(
6.674
Γ
1
0
β
11
β
m
3
/
kg
β
s
2
)
β
(
1.0
Γ
1
0
13
β
kg
)
β
(
30
,
000
β
m
)
3
T
2
=
(6.674Γ10
β11
m
3
/kgβ s
2
)β (1.0Γ10
13
kg)
4Ο
2
β
β (30,000m)
3
2
=
4
2
6.674
Γ
1
0
β
11
β
1.0
Γ
1
0
13
β
(
27
Γ
1
0
9
)
T
2
=
6.674Γ10
β11
β 1.0Γ10
13
4Ο
2
β
β (27Γ10
9
)
2
=
4
2
β
27
Γ
1
0
9
6.674
Γ
1
0
2
T
2
=
6.674Γ10
2
4Ο
2
β 27Γ10
9
β
2
β
4.0732
Γ
1
0
14
T
2
β4.0732Γ10
14
Now, take the square root of both sides to find
T:
β
4.0732
Γ
1
0
14
Tβ
4.0732Γ10
14
β
β
6.3832
Γ
1
0
7
β
seconds
Tβ6.3832Γ10
7
seconds
Now, let's convert the orbital period from seconds to days:
days
=
seconds
86400
T
days
β
=
86400
T
seconds
β
β
days
=
6.3832
Γ
1
0
7
86400
T
days
β
=
86400
6.3832Γ10
7
β
days
β
738.39
β
days
T
days
β
β738.39days
So, the orbital period of the probe around the comet is approximately 738.39 days.
