Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [ Hint: Equation 13.6 does not apply. Use Equation 13.5 and include the potential energy of both Earth and the Sun. Substituting the values for Earth’s mass and radius directly into Equation 13.6, we obtain v e s c = 2 G M R = 2 ( 6.67 × 10 − 11 N ⋅ m 2 / k g 2 ) ( 5.96 × 10 24 k g ) ( 6.37 × 10 6 m ) = 1.12 × 10 4 m / s That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use R = R E S = 1.50 × 10 11 m and M S u m = 1.99 × 10 30 kg . The result is v esc = 4.21 × 10 4 m / s or about 42 km/s. We have 1 2 m v e s c 2 − G M m R = 1 2 m 0 2 − G M m ∞ = 0 Solving for the escape velocity,
Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [ Hint: Equation 13.6 does not apply. Use Equation 13.5 and include the potential energy of both Earth and the Sun. Substituting the values for Earth’s mass and radius directly into Equation 13.6, we obtain v e s c = 2 G M R = 2 ( 6.67 × 10 − 11 N ⋅ m 2 / k g 2 ) ( 5.96 × 10 24 k g ) ( 6.37 × 10 6 m ) = 1.12 × 10 4 m / s That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use R = R E S = 1.50 × 10 11 m and M S u m = 1.99 × 10 30 kg . The result is v esc = 4.21 × 10 4 m / s or about 42 km/s. We have 1 2 m v e s c 2 − G M m R = 1 2 m 0 2 − G M m ∞ = 0 Solving for the escape velocity,
Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [Hint: Equation 13.6 does not apply. Use Equation 13.5 and include the potential energy of both Earth and the Sun.
Substituting the values for Earth’s mass and radius directly into Equation 13.6, we obtain
v
e
s
c
=
2
G
M
R
=
2
(
6.67
×
10
−
11
N
⋅
m
2
/
k
g
2
)
(
5.96
×
10
24
k
g
)
(
6.37
×
10
6
m
)
=
1.12
×
10
4
m
/
s
That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use
R
=
R
E
S
=
1.50
×
10
11
m
and
M
S
u
m
=
1.99
×
10
30
kg
. The result is
v
esc
=
4.21
×
10
4
m
/
s
or about 42 km/s.
We have
1
2
m
v
e
s
c
2
−
G
M
m
R
=
1
2
m
0
2
−
G
M
m
∞
=
0
The star Sirus A has a mass of 2.06 MO and a radius of 1.71 RO, where M0 is the mass of the Sun
(1.988 x 1030 kg) and RO is the radius of the Sun (6.96 x 105 km).
(a) Sketch the gravitational potential of Sirus A, which a hydrogen particle would experience at
distances where r is greater than the radius of Sirus A.
(b) Calculate the gravitational potential energy of the particle-star system when the hydrogen
particle has reached a distance of 10 RO. Note the atomic mass of hydrogen is 1.0079 amu.
Question 3
a. A space craft achieved a maximum speed of 125,000 km/h on its way to photograph Mars. Beyond
what distance from the Sun is this speed sufficient to escape the solar system?
b. Consider a large massive spherical shell object. With all of its mass M is distributed at its radius R
(shown in the figure). Draw a schematic granh of gravitational force and corresponding
Page 1 of 2
gravitational potential energy experienced by another object m at a distance r, where the distance
varies as, r → 0 to r 0o.
Figure 2: Question 3(b) Two masses system
Zero, a hypothetical planet, has a mass of 5.0 * 102^3 kg, a radius of 3.0 * 10^6 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of 5.0 * 107 J, what will be its kinetic energy when it is 4.0 * 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 * 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
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