24P3Q02

2a)

Gravitational pull provides required centripetal force.

\displaystyle \frac{GMm}{{{(R+h)}^{2}}}=m\frac{{{v}^{2}}}{(R+h)}\text{ }\Rightarrow \text{ }v=\sqrt{\frac{GM}{R+h}}

\displaystyle {{E}_{k}}=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \sqrt{\frac{GM}{R+h}} \right)}^{2}}=\frac{GMm}{2(R+h)}

2b)

\displaystyle \begin{aligned}   {{E}_{T}}=KE+GPE &=\frac{1}{2}\frac{GMm}{(R+h)}+(-\frac{GMm}{R+h}) &=-\frac{1}{2}\frac{GMm}{R+h}\end{aligned}

2c)

Gravitational pull provides required centripetal force.

\displaystyle \frac{GMm}{{{(R+h)}^{2}}}=m\frac{{{v}^{2}}}{(R+h)}\text{ }\Rightarrow \text{ }v=\sqrt{\frac{GM}{R+h}}

\displaystyle \begin{aligned}v&=\sqrt{\frac{GM}{R+h}} &=\sqrt{\frac{(6.67\times {{10}^{-11}})(7.35\times {{10}^{22}})}{1.74\times {{10}^{6}}+110000}} &=1620\text{ m }{{\text{s}}^{-1}}\end{aligned}

2di)

P has higher Ek.

Since \displaystyle {{E}_{k}}=\frac{GMm}{2(R+h)}, Ek is higher when h is smaller.

This is because at lower altitude, the satellite must orbit faster in order for the centripetal acceleration to match the stronger gravitational acceleration.

2dii)

P has lower ET.

Since \displaystyle {{E}_{T}}=-\frac{GMm}{R+h}, ET is more negative when h is smaller.

This is because the GPE is more negative at lower altitude.

One thought on “24P3Q02”

  1. Pingback: A-Level 2024 – xmPhysics

Leave a comment