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An asteroid has a mass of 52000 kg semi major axis ? =3.6 AU (1 AU= 1.5 e11 m) with an eccentricity ? =0.5. The mass of the sun is 2.0 e30 kg 3. Using conservation of momentum find the velocity of the asteroid at perihelion. 4. Find the period of the asteroid. 5. Find the velocity of the asteroid at 3.5 AU.

An asteroid has a mass of 52000 kg semi major axis ? =3.6 AU (1 AU= 1.5 e11 m) with an eccentricity ? =0.5. The mass of the sun is 2.0 e30 kg 3. Using conservation of momentum find the velocity of the asteroid at perihelion. 4. Find the period of the asteroid. 5. Find the velocity of the asteroid at 3.5 AU.

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An asteroid has a mass of 52000 kg semi major axis ? =3.6 AU (1 AU= 1.5 e11 m) with an eccentricity ? =0.5. The mass of the sun is 2.0 e30 kg

3. Using conservation of momentum find the velocity of the asteroid at perihelion.
4. Find the period of the asteroid.
5. Find the velocity of the asteroid at 3.5 AU.

It a Is the semi-major of the eliptical orbit of an astronomical object of mass m around the sun, then, the total mechanical energy of the astronomical object is given by GM sm ET = - 2a Here G is the universal gravitational constant and Ms is the mass of the sun. The negative sign indicates that the force between the sun and the astronomical object is attractive. If r is the distance of the object from the sun then the velocity of the object is GMs ( - 4) v = For an elliptical orbit, the distance of the object from the sun at the perihelion is r = a (1 – e) Step 2 Given the mass of the asteroid m = 52 x 103 kg length of the semi-major axis a = 3. 6 AU = 3. 6 × 1. 5 × 10m Mass of the sun Ms = 2 x 1030 kg Universal gravitational constant G = 6. 67428 × 10¬11 N kg2 The eccentricity of the orbit e = 0. 5 a) Therefore the total mechanical energy of the asteroid 6.67428×10-11x2x10³º×52×103 2x3.6x1.5×10'1 = -6. 427 x 1012 J ET = - 2a b) Distance of the asteroid from the sun as the perihelion r = a (1 – e) = 3. 6 × 1. 5 × 10 × (1 – 0. 5)m = 2.7 × 1011 m - - Therefore the velocity of the asteroid at the perihelion GMs ( - ±) = /GM5 (a - GMs (a-e) v=. %D 1 6. 67428 × 10-11 × 2 × 1030 × 2.7x10!1 3.6x1.5×10'' =27232. 089 m/s = 2. 7232 × 104 m/s Step 3 Answer: a. Total energy of the asteroid ET -6. 427 × 1012 J. b. The velocity of the asteroid at the perihelion is v = 2. 7232 × 104 m/s. Privacy - Terms
Transcribed Image Text:It a Is the semi-major of the eliptical orbit of an astronomical object of mass m around the sun, then, the total mechanical energy of the astronomical object is given by GM sm ET = - 2a Here G is the universal gravitational constant and Ms is the mass of the sun. The negative sign indicates that the force between the sun and the astronomical object is attractive. If r is the distance of the object from the sun then the velocity of the object is GMs ( - 4) v = For an elliptical orbit, the distance of the object from the sun at the perihelion is r = a (1 – e) Step 2 Given the mass of the asteroid m = 52 x 103 kg length of the semi-major axis a = 3. 6 AU = 3. 6 × 1. 5 × 10m Mass of the sun Ms = 2 x 1030 kg Universal gravitational constant G = 6. 67428 × 10¬11 N kg2 The eccentricity of the orbit e = 0. 5 a) Therefore the total mechanical energy of the asteroid 6.67428×10-11x2x10³º×52×103 2x3.6x1.5×10'1 = -6. 427 x 1012 J ET = - 2a b) Distance of the asteroid from the sun as the perihelion r = a (1 – e) = 3. 6 × 1. 5 × 10 × (1 – 0. 5)m = 2.7 × 1011 m - - Therefore the velocity of the asteroid at the perihelion GMs ( - ±) = /GM5 (a - GMs (a-e) v=. %D 1 6. 67428 × 10-11 × 2 × 1030 × 2.7x10!1 3.6x1.5×10'' =27232. 089 m/s = 2. 7232 × 104 m/s Step 3 Answer: a. Total energy of the asteroid ET -6. 427 × 1012 J. b. The velocity of the asteroid at the perihelion is v = 2. 7232 × 104 m/s. Privacy - Terms
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