In: Physics

Consider a satellite of mass ms in circular orbit around Earth, a distance h above Earth's surface. Assume the Earth is a sphere with radius Re and mass Me.

(a) As the satellite travels in circular orbit, will its speed increase, decrease, or remain constant? Explain.

(b) The only force acting on the satellite is gravity, so the satellite is in freefall. Why doesn't the satellite get closer to Earth's surface?

(c) Determine the ratio of the force of gravity on the satellite while in orbit to the force of gravity on the satellite while on the surface of Earth. Simplify your answer; no complex fractions! Give your answer in terms of Re and h.

(d) Starting with Newton's second law, determine an equation for each of the following quantities. Give your answers in terms of Me, Re, h, and the universal gravitation constant G.

(i) the freefall acceleration of the satellite,

(ii) The orbital speed of the satellite,

(iii) the period of orbit, which is the time it takes the satellite to complete one orbit around Earth.

(e) Determine the work done by gravity as the satellite moves through one-half of its circular orbit.

(f) Consider a near-Earth orbiting satellite (h=0.10Re) that is inhabited by astronauts.

(i) Determine the ration of the freefall acceleration of an astronaut in this circular orbit to the freefall acceleration of an astronaut on Earth.

(ii) If that astronaut weights 150 pounds on Earth, how large is the force of gravity on her in orbit? Is she "weightless"?

a) In a circular orbit, speed always remains constant. The change is speed is observed in elliptical orbits. In circular orbit, distance is always constant and hence speed, which is in accordance with the law of constant aerial velocity.

b) It is true that the atellites are in freefall, but the rate at which the falls toward earth is eqaul to the rate at which earth curves away from satellite. It is similar to a situation of a dog trying to catch its own tail.

c) Force on surface of earth Fs= (G*Me*Ms)/Re^2

Force in orbit Fo= (G*Me*Ms)/(Re+h)^2

Fo/Fs = [Re/(Re+h)]^2

d)i) F = G*Me*Ms / (Re + h)^2 = Ms*a

a = G*Me / (Re + h)^2

ii) a = V^2 / (Re + h)

V = sqrt[ a*(Re+h)]

V = sqrt [G*Me / (Re+h)]

iii) T = 2*pi*(Re+h) / V

T = [2*pi*(Re+h)] / sqrt [G*Me / (Re+h)]

T = [2*pi*(Re+h)^(3/2)] / sqrt(G*Me)

e) As the force of gravity is directed toward centre of earth and displacement is in perpendicular direction, hence,

work done by gravity is zero

f)i) acceleration in orbit = G*Me / (Re + 0.1Re)^2 = G*Me / (1.21*Re^2)

acceleration on surface = G*Me / Re^2

Ratio = [G*Me / (1.21*Re^2)] / [ G*Me / Re^2]

= 0.826

ii) in the orbit the wieght will reduce by the ratio calculayed in above part i.e

W = 150*0.826 = 123.96 lb

An Earth satellite moves in a circular orbit 924 km above
Earth's surface with a period of 103.3 min. What are
(a) the speed and (b) the
magnitude of the centripetal acceleration of the satellite?

A spy satellite is in circular orbit around Earth. It makes one
revolution in 6.10 h. Mass of Earth is 5.974 × 1024 kg,
radius of Earth is 6371 km and Gravitational constant G is = 6.674
× 10−11 N·m2/kg2.
1. How high above Earth’s surface is the satellite? _____km
2.What is the satellite’s acceleration? _____ m/s^2

A satellite is placed in an elongated elliptical (not circular)
orbit around the Earth. At the point in its orbit where it is
closest to the Earth, it is a distance of 1.00 × 10^6
m from the surface (not the center) of the Earth,
and is moving at a velocity of 5.14 km/s. At the
point in its orbit when it is furthest from the Earth it is a
distance of 2.00×10^6 m from the surface of the
Earth....

Two satellites are in circular orbits around the earth. The
orbit for satellite A is at a height of 555 km above the earth’s
surface, while that for satellite B is at a height of 778 km. Find
the orbital speed for (a) satellite A and (b) satellite B.

A satellite moves in orbit about the Earth at a height of 9500
km above Earth's surface. Find a) the velocity of the satellite; b)
it's period of rotation; c) the magnitude of the Earth's
gravitational force on the satellite if it's mass is 2500 kg; d) if
the satellite's velocity is increased by 20% at what height must it
now orbit the Earth?
Explain D step by step please.

A geostationary satellite is in a circular orbit around the
Earth. What is the linear speed of the satellite?

A satellite in a circular orbit around the earth with a radius
1.011 times the mean radius of the earth is hit by an incoming
meteorite. A large fragment (m = 81.0 kg) is ejected in the
backwards direction so that it is stationary with respect to the
earth and falls directly to the ground. Its speed just before it
hits the ground is 361.0 m/s. Find the total work done by gravity
on the satellite fragment. RE 6.37·103 km;...

A 590-kg satellite is in a circular orbit about Earth at a
height above Earth equal to Earth's mean radius.
(a) Find the satellite's orbital speed.
m/s
(b) Find the period of its revolution.
h
(c) Find the gravitational force acting on it.
N

We want to place a satellite moving in a circular orbit 500 km above the earth's surface in a circular orbit 1000 km above the earth's surface. To do this, first give the satellite a speed increase of ikenv while it is in the 1st green orbit and place it in the 2nd yellow elliptical orbit, and then, when the elliptical orbit reaches a distance of 1000 km from the ground, it gives it a second Δv 'velocity and allows...

A satellite with Mass m is in orbit with a constant radius
around the earth r0 (RE=6370km, Mass
ME = 5,98*1024kg)
a) Show that the satellite moves with uniform circular motion
and calculate the velocity v0 in dependance of G,M
E and R E .
b) At which height h above the earth's surface is the
geostationary orbit found? Which linear velocity does a satellite
have at this height?
c) Compare this to the linear velocity on earth's surface as...

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