An object is released from rest when it is a height h above the surface of a pla

Question

An object is released from rest when it is a height h above the surface of a planet of mass M and radius R. What is the speed of the object just before striking the surface of the planet? Neglect any air resistance. Let h = 4.0 Times 10^6 m, R = 5.0 Times 0^6 m, and M = 4.0 Times 10^24 kg. A satellite is placed in a geosynchronous orbit. In this equatorial orbit with a period of 24 hours, the satellite hovers over one point on the earth's equator. What is the height of the satellite above the planet's surface? What is the gravitational field magnitude at the location of the satellite from the previous problem?

Explanation / Answer

ans 11

About 42,164 km (26,199 mi) (measured from the center of the Earth). A satellite in such an orbit is at an altitude of approximately 35,786 km (22,236 mi) above mean sea level.

ans 12

Geosynchronous satellites orbit the Earth at an altitude of about 3.5786km. Given that the Earth's radius is 6.38*106 m and its mass is 5.97*1024 kg

so by formula

since this is a problem asking about the earth's gravity, you want the distance from the center of the earth;

now, g=GM/r2 where G is the newtonian gravity constant M is the mass of the earth, and r is the orbital distance

the easiest way to find the local accel due to gravity is just take ratios:

g(in orbit)/g(surface)=[R(surface)/altitude...

we know R(surface)=6.38x106m and altitude=4.2164x107m

so, g(orbit)/g(surf)=(6.38*106/4.2164*107)2 = 0.022896

g(orbit)=9.8m/s/s * 0.022896=0.22437m/s2

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