Question #6cea0

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)]#