Answer :
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
⋅
⋅
3
T
2
=
G⋅M
4π
2
⋅r
3
Where:
T is the orbital period (in seconds)
G is the gravitational constant (
6.674
×
1
0
−
11
m
3
/
kg
⋅
s
2
6.674×10
−11
m
3
/kg⋅s
2
)
M is the mass of the comet (
1.0
×
1
0
13
kg
1.0×10
13
kg)
r is the distance between the probe and the center of the comet (
30
km
=
30
,
000
m
30km=30,000m)
First, let's convert the distance from kilometers to meters:
=
30
,
000
m
r=30,000m
Now, we can plug the values into the equation:
2
=
4
2
(
6.674
×
1
0
−
11
m
3
/
kg
⋅
s
2
)
⋅
(
1.0
×
1
0
13
kg
)
⋅
(
30
,
000
m
)
3
T
2
=
(6.674×10
−11
m
3
/kg⋅s
2
)⋅(1.0×10
13
kg)
4π
2
⋅(30,000m)
3
2
=
4
2
6.674
×
1
0
−
11
⋅
1.0
×
1
0
13
⋅
(
27
×
1
0
9
)
T
2
=
6.674×10
−11
⋅1.0×10
13
4π
2
⋅(27×10
9
)
2
=
4
2
⋅
27
×
1
0
9
6.674
×
1
0
2
T
2
=
6.674×10
2
4π
2
⋅27×10
9
2
≈
4.0732
×
1
0
14
T
2
≈4.0732×10
14
Now, take the square root of both sides to find
T:
≈
4.0732
×
1
0
14
T≈
4.0732×10
14
≈
6.3832
×
1
0
7
seconds
T≈6.3832×10
7
seconds
Now, let's convert the orbital period from seconds to days:
days
=
seconds
86400
T
days
=
86400
T
seconds
days
=
6.3832
×
1
0
7
86400
T
days
=
86400
6.3832×10
7
days
≈
738.39
days
T
days
≈738.39days
So, the orbital period of the probe around the comet is approximately 738.39 days.
