A satellite of mass m, originally on the surface of the...

A satellite of mass $m,$ originally on the surface of the Earth, is placed into Earth orbit at an altitude $h$ . (a) With a circular orbit, how long does the satellite take to complete one orbit? (b) What is the satellite's speed? (c) What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet's daily rotation. At what location on the Earth's sur- face and in what direction should the satellite be launched to minimize the required energy investment? Represent the mass and radius of the Earth as $M_{E}$ and $R_{E}$ .

A satellite of mass $m,$ originally on the surface of the Earth, is placed into Earth orbit at an altitude $h$ . (a) With a circular orbit, how long does the satellite take to complete one orbit? (b) What is the satellite's speed? (c) What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet's daily rotation. At what location on the Earth's sur-
face and in what direction should the satellite be launched to minimize the required energy investment? Represent the mass and radius of the Earth as $M_{E}$ and $R_{E}$ .

Key Concepts

Energy in Orbital Mechanics

Orbit insertion and maintenance involve changes in both gravitational potential energy and kinetic energy. The minimum energy input required to place a satellite into orbit is determined by the difference between the energy state on the surface and the orbital energy, which includes both the kinetic energy associated with its orbital speed and the change in gravitational potential energy. Understanding these energy considerations is key to calculating the efficiency and requirements of orbital insertion.

Effect of Earth's Rotation

The rotation of the Earth imparts an initial tangential velocity to objects on its surface. When launching a satellite, this initial velocity can reduce the additional energy required to achieve orbital speed relative to an inertial frame. Accounting for Earth's rotation is therefore important for accurately determining the minimum energy input necessary to place a satellite into orbit.

Newton's Law of Universal Gravitation

This law states that every point mass attracts every other point mass with a force along the line intersecting both points. The magnitude of the force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. It is fundamental in calculating the gravitational force acting on satellites and forms the basis for understanding orbital dynamics.

Circular Orbital Motion

In a stable circular orbit, the gravitational force provides the necessary centripetal force to keep a satellite in its path. By equating the gravitational force to the centripetal force, one can derive important relationships such as the orbital speed and period. These relationships are critical for determining how long a satellite takes to complete one orbit and the speed required to maintain such an orbit at a given altitude.

Transcript

00:01 Looking at a satellite that's orbiting the earth.

00:03 Okay, and it's launched from the earth's surface to a height h.

00:09 And i want to find, for a satellite of mass m, i want to find its orbital period, so the time it takes to complete one full rotation around the earth.

00:20 I want to find its velocity.

00:24 Okay.

00:26 And i want to find how much energy we must input to the rocket system to, put it into orbit at this height h.

00:34 Okay, so my change in energy from the earth's surface where it was launched from to its final orbiting position of h meters above the earth's surface.

00:48 And i'm going to be finding all of these quantities in terms of h, the satellite's orbital height, mass of the satellite, the mass of the earth, the gravitational constant, and the radius of the earth.

01:05 Okay, so let's start by finding our orbital period.

01:10 All right.

01:12 And i know that for orbital motion, my velocity equals 2 pi radius over t.

01:23 And here we're looking for my velocity when i'm in orbit.

01:26 So the radius is radius of the earth plus my orbital height.

01:31 Know that for circular motion, acceleration equals v squared over r.

01:39 So subbing in our expression for velocity here, i get 4 pi squared over t squared, r .e plus h squared over r .e plus h.

01:58 You'll see that one cancels out.

02:01 And finally, i have from newton's second law, mass times acceleration of a body.

02:08 The body in this case is our satellite.

02:11 Equals the sum of all forces acting on that body.

02:15 The only force we have is the force due to gravity between the satellite and the earth.

02:21 And we know that that force equals m, m, where m is the mass of the earth, m is the mass of our satellite.

02:29 G, gravitational constant, over the distance between the two bodies squared, so radius of the earth plus h squared.

02:41 It will see that the m drops out on both sides, and i'll substitute the value we just found for acceleration here.

02:51 So that gives me 4 pi squared, r .e.

02:57 Plus h over t, squared.

03:01 Squared equals m -e -g over r -e -plus -h squared.

03:13 And solving this for t, i get t -squared equals 4 pi squared, r -e -plus -h -cubed, all over mass of earth times gravitational.

03:33 Constant or t equals 2 pi r e plus h to the three halves over the square root of mass of the earth times gravitational constant now we can easily find an expression for velocity remember velocity equals 2 pi r over t right so we have 2 pi r r e plus h over t and with our expression we just found for t here, that is over 2 pi, r .e plus h to the three halves times square root of mass of earth times g.

04:31 It can simplify this a bit further.

04:33 We can get rid of the two pi.

04:35 We can get rid of that...

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