A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is (¼)th of its value of the surface of the planet. If the escape velocity from the planet is Vesc = v VN, then the value of N is (ignore energy loss due to atmosphere).
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- A spaceship with m = 1.00 ✕ 104 kg is in a circular orbit around the Earth, h = 800 km above its surface. The ship's captain fires the engines in a direction tangent to the orbit, and the spaceship assumes an elliptical orbit around the Earth with an apogee of d = 3.00 ✕ 104 km, measured from the Earth's center. How much energy (in J) must be used from the fuel to achieve this orbit? (Assume that all the fuel energy goes into increasing the orbital energy and that the perigee distance is equal to the initial radius.)Consider an object that is in an elliptical orbit with semimajor axis a = 7.9×106 m about a planet of mass M = 1.0×1023 kg. (a) What is the speed of the object when it is closest to the planet at r = a/4? (b) What is the speed of the object when it is furthest from the planet?The escape velocity from a massive object is the speed needed to reach an infinite distance from it and have just slowed to a stop, that is, to have just enough kinetic energy to climb out of the gravitational potential well and have none left. You can find the escape velocity by equating the total kinetic and gravitational potential energy to zero E = = muesc - GmM/r=0 Vesc = √2GM/r where G is Newton's constant of gravitation, M is the mass of the object from which the escape is happening, and r is its radius. This is physics you have seen in the first part of the course, and you should be able to use it to find an escape velocity from any planet or satellite. For the Earth, for example the escape velocity is about 11.2 km/s, and for the Moon it is 2.38 km/s. A very important point about escape velocity: it does not depend on what is escaping. A spaceship or a molecule must have this velocity or more away from the center of the planet to be free of its gravity, 1. In the atmosphere of…
- (a) What is the escape speed on a spherical asteroid whose radius is 502 km and whose gravitational acceleration at the surface is 0.870 m/s? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 647 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 665.5 km above the surface? (a) Number i Units (b) Number Units (c) Number UnitsA space capsule of mass 505 kg is at rest 8.00 x 10' m from the center of the Earth. When it has fallen 5.00 x 10° m closer to the Earth, finu le (a) What is the change in the system's gravitational potential energy? (b) Find the speed of the satellite at that point. m/sA planet has a radius of 4.00 x 10^6 m, and rotates so rapidly that an object on the equator feels only 10% of the weight that it feels at the poles. What is the speed of an object at the equator?
- A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5 x 1010 m (inside the orbit of Mercury), at which point its speed is 9.6 x 10* m/s. Its farthest distance from the Sun is beyond the orbit of Pluto. What is its speed when it is 6 x 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.) speed = m/s Additional Materials eBook O Show My Work (Optional)A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.5 x 1010 m (inside the orbit of Mercury), at which point its speed is 9.2 x 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 x 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.) speed = 690 X m/s Additional Materials eBookAn object is projected from the surface of the earth with a speed of 2.72×104 m/s. What is its speed when it is very far from the earth? (Neglect air resistance.)
- A 3500-kg spaceship is in a circular orbit 220 km above the surface of Earth. It needs to be moved into a higher circular orbit of 390 km to link up with the space station at that altitude. In this problem you can take the mass of the Earth to be 5.97 × 1024 kg.How much work, in joules, do the spaceship’s engines have to perform to move to the higher orbit? Ignore any change of mass due to fuel consumption.Two masses m, = 100 kg and m, = 8100 kg are held 1 m apart. (a) At what point on the line joining them is the gravitational field equal to zero? Find the gravi- tational potential at that point. (b) Find the gravitational potential energy of the system. Given G = 6.67 × 10-" Nm? kg.One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R, and mass 4M. If M = 4.07 × 1024 kg and R = 5.79 x 106 m, what is the gravitational acceleration of a particle at points (a) R and (b) 3R from the center of the planet? (a) Number i Units (b) Number i Units