Gravitational Fields
Equations
\[g=\frac{GM_{1}}{r^2}$$
$$F=\frac{GM{1}M_{2}}{r^2}$$
$$g=\frac{F}{m}$$
$$w=mg$$
$$E_{g}=mgh\]
Gravity and radius relationship
\[\frac{g_{1}}{g_{2}}=\frac{r_{s_{2}}}{r_{A_{2}}}\]
At a point \(x\), g is 0, due to it being in the middle of two planets of mass M and 4M .
- \(g_{p}=g_{a}\)
- \(r\) is the distance between the two planets
- \(a\) is the distance from planet \(P\) to point \(x\)
- \(\frac{GM_{p}}{a^2}=\frac{GM_{Q}}{(r-a)^2}\)
- \(M_{Q}=4M_{p}\)
- \(\frac{GM_{p}}{a^2}=\frac{4GM_{p}}{(r-a)^2}\)
- \(\frac{1}{a^2}=\frac{4}{(r-a)^2}\)
- \(a^2=\frac{(r-a)^2}{4}\)
- \(4a^2=(r-a)^2\)
- \(2a=r-a\)
- \(3a=r\)
Calculate the gravitational field strength on a planet...
a) same mass and twice radius of earth.
- \(g=\frac{GM}{r^2}\)
- \(r_{p}=2r_{\text{ earth}}\)
- \(m_{p}=m_{\text{ earth}}\)
- \(g_{p}=\frac{GM}{(2r)^2}=\frac{GM}{4r^2}\)
- \(g_{p}=\frac{1}{4} \text{ }g_{\text{ earth}}\)
b) same mass and twice density
- \(p_{p}=2p\)
- \(p=\frac{m}{v}\)
- \(v=\frac{4\pi r^3}{3}\)
- \(m=\frac{4\pi r^3}{3}p\)
- \(M_{p}=\frac{4\pi r_{p}^32p}{3}\)
- \(r^3=\frac{3M}{4\pi p}\)
- \(r_{p}=\frac{3M}{4\pi2p}\)
- \(r^3=2r^3_{p}\)
- \(r=(2)^{\frac{1}{3}}r_{p}\)
- \(\frac{g_{p}}{g}=\frac{\cancel{ M }r^2}{\cancel{ M }r_{p}^2}\)
- \(\frac{g_{p}}{9.81}=\frac{2^{\frac{2}{3}}r_{p}^2}{r_{p}^2}\)
- \(g_{p}=9.81\times 2^{\frac{2}{3}}\)
- \(=15.57\)
- \(=16N kg^{-1}\)