- $6\cdot10^{24}\,\text{kg}$
- $6\cdot10^{27}\,\text{kg}$
- $6\cdot10^{30}\,\text{kg}$
- $6\cdot10^{33}\,\text{kg}$
- $6\cdot10^{36}\,\text{kg}$

Solution :

Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.

**Choice 1 is the answer.**

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

- $6\cdot10^{24}\,\text{kg}$
- $6\cdot10^{27}\,\text{kg}$
- $6\cdot10^{30}\,\text{kg}$
- $6\cdot10^{33}\,\text{kg}$
- $6\cdot10^{36}\,\text{kg}$

Solution :

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

Subscribe to:
Post Comments (Atom)

## 1 comment:

## Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

Knowing little things like that I feel help arrive at an answer faster than doing the calculation.

Post a Comment