Question

How does mass affect orbital speed?

Answer 1

Here in this question, we have Assumed that we are talking about the mass of the satellite and not the mass of the body being orbited. The relationship between the time period of rotation and the distance between the sun is described by Kepler's third law.

Complete Step By Step Answer:
According to the Kepler's 3rd Law of Planetary Motion:
${T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{GM}}$
Where,
$T$ is known as the period of the orbit.
The orbital speed determines the radius of the orbit.
The constants in that equation are the terms $\dfrac{{4{\pi ^2}}}{G}$ . I'll presume you want to compare masses without changing the radius, therefore the full formula $\dfrac{{4{\pi ^2}{r^3}}}{G}$ will be treated as constant.
It's unclear which mass you're referring to. There is the mass of the satellite to consider, as well as the mass of the body being orbited.
I'm going to assume you're inquiring about the satellite's mass. The mass of the satellite is not included in the formula $\dfrac{{4{\pi ^2}{r^3}}}{{GM}}$ . As a result, we can deduce that the satellite's mass has no bearing on its orbital speed.
I'm assuming you're inquiring about the mass of the object being orbited. The $M$ in the equation $\dfrac{{4{\pi ^2}{r^3}}}{{GM}}$ stands for the mass of that body. As a result, as $M$ grows, $T$ decreases, indicating that the speed has increased.

Note:
The velocity at which an object's kinetic energy approaches the fundamental of all likely energy out to infinity, implying that the object might totally escape the gravity of a body, is known as escape velocity. On the other side, orbital velocity is determined by the specific orbital route; however, for a roundabout circle, it deviates from the velocity at the distance separated by the square root of 2. As a result, the object is falling in a circular motion around the centre body rather than vanishing into infinity.