# How to Launch a Satellite with Apogee 2.5 Times the Planet's Radius?

• Zatman
In summary, the conversation discusses the use of conservation of energy and angular momentum to calculate the necessary velocity for launching a satellite into orbit with a specific apogee. It is assumed that the planet is spherical, not rotating, and atmospheric effects can be ignored. After realizing and correcting a mistake in the energy equation, the necessary velocity is determined to be v_0 = √(5GM/4R).
Zatman

## Homework Statement

It is required to put a satellite of mass m into orbit with apogee 2.5 times the radius of the planet of mass M. The satellite is to be launched from the surface with speed v0 at an angle of 30° to the local vertical.

Use conservation of energy and angular momentum to show that

$v_0^2 = \frac{5GM}{4R}$

assuming that the planet is spherical, not rotating and atmospheric effects can be ignored.

## Homework Equations

$L = mr^2\dot{\theta}$

$V = -\frac{GMm}{r}$

where r is the distance from the centre of the planet to the satellite's position at time t.

## The Attempt at a Solution

At t = 0, the angular momentum is L = mRv0sin(30) = (1/2)mRv0. Therefore:

$\dot{\theta} = \frac{L}{mr^2} = \frac{Rv_0}{2r^2}$

Total energy at time t is:

$E = \frac{1}{2}mv^2 - \frac{GMm}{r}$

$E = \frac{1}{2}mr^2\dot{\theta}^2 - \frac{GMm}{r}$

$E = \frac{mR^2v_0^2}{8r^2} -\frac{GMm}{r}$

I don't really know what to do with these equations. At the apogee, dr/dt = 0, so I could differentiate the E equation, but then I just get 0 = 0. Also I notice that E should be the same for all r as well as t, but clearly E(r) is not constant, so is the energy equation wrong?

Any help would be greatly appreciated.

The energy equation is wrong, because you neglected the radial velocity.

1 person
Well, don't I feel silly now.

So including the radial velocity gives a term with dr/dt. I set the energy equation equal to the initial energy at t=0 {i.e. (1/2)mv02 - GMm/R}. Then at the apogee, r=2.5R and dr/dt = 0, solve for v_0 which gives the required result.

Thank you for pointing out my silly error.

## 1. How do you put a satellite into orbit?

To put a satellite into orbit, it needs to be launched into space on top of a rocket. The rocket will provide the necessary thrust and speed to reach the desired orbit. Once in space, the satellite will use its own propulsion systems to fine-tune its orbit and maintain its position.

## 2. What factors determine the orbit of a satellite?

The orbit of a satellite is determined by several factors, including the initial launch trajectory, the amount of thrust provided by the rocket, and the gravitational pull of the Earth and other celestial bodies. The desired purpose of the satellite, such as communication or weather monitoring, also plays a role in determining its orbit.

## 3. How long does it take to put a satellite into orbit?

The time it takes to put a satellite into orbit varies depending on the type of orbit and the specific launch mission. In general, it can take several minutes to reach the desired orbit, but it may take days or even weeks for the satellite to fully adjust and stabilize in its orbit.

## 4. How high up is a satellite in orbit?

The height of a satellite in orbit depends on the type of orbit it is in. For example, a low Earth orbit can range from 160 to 2,000 kilometers above the Earth's surface, while a geostationary orbit is typically around 36,000 kilometers above the Earth. The height is determined based on the desired purpose and function of the satellite.

## 5. Can a satellite fall out of orbit?

Yes, a satellite can fall out of orbit due to various factors such as atmospheric drag, gravitational pull from other celestial bodies, or malfunction of its propulsion systems. When this happens, the satellite will eventually re-enter Earth's atmosphere and burn up, unless it is redirected to a different orbit or sent farther into space.

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