Question

Question #6cea0

Answer 1

I think it's `["M"^0"L T"^(-1)]`.

Explanation:

Dimensional formulas use mass, length, and time to express physical quantities as a propduct of powers of fundamental quantities.

Since velocity is defined as distance divided by time, you can say that

`"velocity" = "distance"/"time" = "meters"/"seconds" = "m"/"s" = "m s"^(-1)`

This means that the dimensional formula for velocity will have

`["M"^(0) "L T"^(-1)] <=> ["L T"^(-1)]`

This tells you that

• you have no units of mass in equation, since `"M"` is raised to the power of zero;
• you have units of length, meters, raised to the power of 1;
• you have units of time, seconds, raised to the power of negative 1;

So, take the equation for escape velocity and test to see if these are the units that you're left with for the dimensional formula.

`v_"esc" = sqrt((2 * G * M)/r)`, where

`G` - the universal gravitational constant, equal to `6.67 * 10^(-11)"m"^(3)"kg"^(-1)"s"^(-2)`
`M` - the mass of the body, expressed in `"kg"`;
`r` - the distance from the center of the body's mass to a point in space.

If you use units, the equation will look like this

`v_"esc" = sqrt( ((2 * "m"^(3)cancel("kg"^(-1))"s"^(-2) * cancel("kg"))/"m")`

`v_"esc" = sqrt( (2 * "m"^(2) * "s"^(-2))`

As far as units are concerned, this will give you

`v_"esc" = "m" * "s"^(-1)`

As you can see, this confirms the initial dimensional formula, since you have meters multiplied by seconds to the power of negative 1. Therefore, the dimensional formula for escape velocity is

`["M"^(0)"L T"^(-1)] <=> ["L T"^(-1)]`