Two satellites are launched at a distance R from a planet of negligible radius. Both satellites are
launched in the tangential direction. The first satellite launches correctly at a speed v0 and enters
a circular orbit. The second satellite, however, is launched at a speed
1
2
v0. What is the minimum
distance between the second satellite and the planet over the course of its orbit?
@drwls it doesn’t matter that the second satellite can’t maintain a circular orbit, you want to find when the second satellite is closest to the planet in its (elliptical) orbit.
To find the minimum distance between the second satellite and the planet over the course of its orbit, we can use the concept of angular momentum conservation.
Let's assume the mass of the planet is M and the mass of the satellite is m.
For the first satellite, launched correctly with a speed v0, it enters a circular orbit at radius R. The centripetal force on the satellite is provided by the gravitational force between the planet and the satellite:
mv0^2 / R = G * M * m / R^2
For the second satellite, launched with a speed of 1/2 v0, we need to find the minimum distance it reaches from the planet.
In order to find the minimum distance, we need to find the radius at which the gravitational force on the satellite due to the planet is equal to the centrifugal force required to keep the satellite in orbit.
Let's assume this minimum distance is r. The centripetal force on the second satellite at this minimum distance r is provided by the gravitational force between the planet and the satellite:
m(1/2v0)^2 / r = G * M * m / r^2
Simplifying the equation, we get:
v0^2 / 4 = G * M / r
Next, we can equate the expressions for centrifugal force in terms of radius R and radius r:
mv0^2 / R = m(1/2v0)^2 / r
Simplifying this equation, we get:
2R = r
So, the minimum distance between the second satellite and the planet is equal to twice the original radius of the circular orbit, which is R. Therefore, the minimum distance between the second satellite and the planet is 2R.
To determine the minimum distance between the second satellite and the planet over the course of its orbit, we need to consider the concept of a Hohmann transfer orbit.
The Hohmann transfer orbit is an elliptical path that transfers a satellite or spacecraft between two circular orbits. In this case, we can use it to determine the minimum distance between the second satellite and the planet.
To calculate the minimum distance, we first need to determine the parameters of the elliptical orbit followed by the second satellite. The semi-major axis (a) of this elliptical orbit can be found using the vis-viva equation:
v^2 = GM * ((2 / r) - (1 / a))
Where:
- v is the speed of the satellite in the orbit (1/2 v0 in this case)
- G is the gravitational constant
- M is the mass of the planet
- r is the distance between the satellite and the planet (R)
Rearranging the equation, we can solve for the semi-major axis:
a = (r / (2 * r - v^2 * r / GM))
Now that we have the semi-major axis (a), we can determine the minimum distance from the planet to the satellite. At the closest point, called the periapsis or the perigee, the distance is equal to:
distance = a * (1 - e)
Where:
- e is the eccentricity of the orbit
Since the satellite is launched in the tangential direction, the eccentricity of its orbit is 1 (e = 1).
Therefore, the minimum distance (distance) can be calculated as:
distance = a * (1 - 1)
distance = a * 0
distance = 0
Hence, the minimum distance between the second satellite and the planet over the course of its orbit is 0.
Where you wrote:
1
2
v0,
is that supposed to be (v0/2) ?
If so, the second satellite does not have sufficient kinetic energy to maintain a circular orbit at distance R.