The geostationary orbit is a unique circular orbit above the equator of a planet, for which the orbital period is equal to the rotation period of the planet. For Earth, the geostationary orbit is 4.22 x 104 km from the centre of the Earth. The Earth rotates in 23 hours and 56 minutes. a) A failed satellite, whose apogee distance is at Earth's geostationary orbit, has only half the required orbital speed of a geostationary satellite at apogee. Calculate the perigee distance of the orbit of this failed satellite. b) If another planet has twice the size of Earth and the same average density as Earth, while rotating in the same amount of time, calculate the radius of the geostationary orbit for this planet.

The geostationary orbit is a unique circular orbit above the equator of a planet, for which the orbital period is equal to the rotation period of the planet. For Earth, the geostationary orbit is 4.22 x 104 km from the centre of the Earth. The Earth rotates in 23 hours and 56 minutes. a) A failed satellite, whose apogee distance is at Earth's geostationary orbit, has only half the required orbital speed of a geostationary satellite at apogee. Calculate the perigee distance of the orbit of this failed satellite. b) If another planet has twice the size of Earth and the same average density as Earth, while rotating in the same amount of time, calculate the radius of the geostationary orbit for this planet.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 182 km? m/s (b) What is the period of revolution? min
A long-distance runner kept up a pace of 8.0 minutes per 1 km for a total of 6530 seconds. How many miles did the runner travel? Assume that 1 mile is equivalent to 1.61 km. Answer isn't 8.447 8.45 4,800 8.248 13.6 6,530 8.5
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.000 × 105 km and the planet has a radius of 3375 km and a mass of 4.75 x 1022 kg, how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67 × 10-¹¹ N·m²/kg². T = days
A synchronous satellite, which always remains above the same point on a planet's equator, is put in circular orbit around Jupiter so that scientists can study a surface feature. Jupiter rotates once every 9.84 h. Use the data of this table to find the altitude of the satellite.
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.270 × 105 km and the planet has a radius of 4.200 × 103 km and a mass of 7.75 × 1022 kg, how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67 × 10-11(Nm2)/(kg2).
You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.50×107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.50×1011 m from its star with a period of 402 earth days. Once on the surface you find that the free-fall acceleration is 13.1m/s2. What is the mass of the planet? Express your answer with the appropriate units. What is the mass of the star? Express your answer with the appropriate units.
A 342 kg satellite currently orbits the Earth in a circle with radius 6.51x10^7m. The satellite must be moved to a new circular orbit of radius 8.22x10^7m. Calculate the additional total mechanical energy that the satellite requires. The mass of the Earth is 5.97x10^24kg G is 6.67x10^-11Nm^2/kg^2
A satellite is in a circular orbit around the Earth at an altitude of 3.82 x 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 x 106 m, and the mass of the Earth is 5.98 x 1024 kg.) h (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite. m/s2 toward the center of the earth
A satellite is in a circular orbit around the Earth at an altitude of 2.62 × 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 x 106 m, and the mass of the Earth is 5.98 x 1024 kg.) h (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite. m/s² toward the center of the earth
A satellite is in a circular orbit around the Earth at an altitude of 3.94 x 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 x 106 m, and the mass of the Earth is 5.98 x 1024 kg.) (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite. m/s² toward the center of the earth
The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5x103 km/h. Knowing that the mean distance from the center of the sun to the center of Venus is 108.5?1021 km and that the radius of the sun is 695.5x103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the sun.
A 639-kg satellite is in a circular orbit about Earth at a height ℎ=1.09e7h=1.09e7 m above the Earth's surface. Find (a) the gravitational force acting on the satellite, (b) the satellite's speed (magnitude of its velocity, notnot its angular velocity), and (c) the period of its revolution. Caution: The radius of the satellite's orbit is not just its height above the Earth's surface. It also includes the radius of the Earth. The mass of the Earth is 5.98×10245.98×1024 kg, and the radius of the Earth is 6.38×1066.38×106 m.