Answers: Here are the answers for each part of the problem:
(a) The hyperbolic excess speed (v_inf) is approximately 9.76 km/s.
(b) The radius (r) when the true anomaly is 100° is approximately 48,497 km.
(c) When the true anomaly is 100°:
- The radial component of the velocity (v_r) is approximately 3.36 km/s.
- The transverse component of the velocity (v_⊥) is approximately 10.6 km/s.
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Step-by-step explanation:
To solve this problem, we'll break it down into three parts.
(a) Calculate the hyperbolic excess speed (km/s)
First, we need to calculate the escape speed (v_esc) at perigee. We use the formula:
v_esc = √(2 * GM / r)
where G is the gravitational constant (6.674 × 10^(-11) m^3 kg^(-1) s^(-2)), M is the mass of Earth (5.972 × 10^24 kg), and r is the distance from the center of the Earth to perigee (r = Earth's radius + perigee altitude = 6371 km + 300 km = 6671 km, converted to meters).
v_esc = √(2 * 6.674 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg / (6,671,000 m))
v_esc ≈ 11.18 km/s
Now, we can find the hyperbolic excess speed (v_inf) using the formula:
v_inf = √(v_perigee^2 - v_esc^2)
where v_perigee is the given perigee speed (15 km/s).
v_inf = √((15 km/s)^2 - (11.18 km/s)^2)
v_inf ≈ 9.76 km/s
(a) The hyperbolic excess speed is approximately 9.76 km/s.
(b) Find the radius (km) when the true anomaly is 100°.
We'll use the equation for the polar equation of a conic section in polar coordinates:
r = (a * (1 - e^2)) / (1 + e * cos(θ))
where r is the radius (distance from the central body), a is the semi-major axis, e is the eccentricity, and θ is the true anomaly. However, we first need to determine the eccentricity and semi-major axis.
We can find the eccentricity (e) using the formula:
e = 1 + (v_inf^2 * r_perigee) / (GM)
e = 1 + ((9.76 km/s)^2 * 6,671,000 m) / (6.674 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg)
e ≈ 1.736
Since this is a hyperbolic trajectory, the semi-major axis (a) will be negative. We can use the following formula to find a:
a = -GM / (2 * v_inf^2)
a = -6.674 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg / (2 * (9.76 km/s)^2)
a ≈ -3,437,000 m (or -3,437 km)
Now, we can find the radius (r) when the true anomaly (θ) is 100°:
r = (-3,437 km * (1 - 1.736^2)) / (1 + 1.736 * cos(100°))
r ≈ 48,497 km
(b) The radius when the true anomaly is 100° is approximately 48,497 km.
(c) Find v_r and v_⊥ (km/s) when the true anomaly is 100°.
We need to find the radial (v_r) and transverse (v_⊥) components of the velocitywhen the true anomaly is 100°. We can use the following equations:
v_r = (GM / h) * e * sin(θ)
v_⊥ = (GM / h) * (1 + e * cos(θ))
where h is the specific angular momentum, GM is the product of the gravitational constant and Earth's mass, e is the eccentricity, and θ is the true anomaly.
First, we need to find the specific angular momentum (h). We can use the formula:
h = r_perigee * v_perigee
h = 6,671,000 m * 15,000 m/s
h ≈ 100,065,000,000 m^2/s
Now, we can find v_r and v_⊥:
v_r = (6.674 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg / 100,065,000,000 m^2/s) * 1.736 * sin(100°)
v_r ≈ 3,360 m/s (or 3.36 km/s)
v_⊥ = (6.674 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg / 100,065,000,000 m^2/s) * (1 + 1.736 * cos(100°))
v_⊥ ≈ 10,600 m/s (or 10.6 km/s)
(c) When the true anomaly is 100°, the radial component of the velocity (v_r) is approximately 3.36 km/s, and the transverse component of the velocity (v_⊥) is approximately 10.6 km/s.