Answer
See the detailed answer below.
Work Step by Step
a) The speed of a point on the equator of the given neutron star is given by
$$v=\dfrac{2\pi R_{star}}{T}$$
$$v=\dfrac{2\pi (10\times 10^3)}{1}=\color{red}{\bf 6.28\times 10^4}\;\rm m/s$$
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b) According to Newton's gravitational law, the free-fall acceleration at the surface of this star is given by
$$g_{\rm surface}=\dfrac{GM_{star}}{R^2_{star}}$$
Plugging the known;
$$g_{\rm surface}=\dfrac{(6.67\times 10^{-11})(1.99\times 10^{30})}{ (10\times 10^3)^2}=\color{red}{\bf 1.33\times 10^{12}}\;\rm m/s^2$$
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c) The weight of the 1-kg mass on the surface of our star is given by
$$W=mg_{\rm surface}=1(1.33\times 10^{12})$$
$$W=\color{red}{\bf 1.33\times 10^{12}}\;\rm N$$
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d) To find the number of revolutions per minute, we need first to find the period of one revolution.
$$F_{G}=\dfrac{GM_{star} \color{red}{\bf\not}m_{\rm satellite}}{R_{orbit}^2}= \color{red}{\bf\not}m_{\rm satellite}a_r$$
where $R_{orbit}=R_{star}+h$ and $a_r=v^2/R_{orbit}$
$$ \dfrac{GM_{star} }{(R_{star}+h)^2}= \dfrac{v^2}{(R_{star}+h)}$$
where $v=2\pi R_{orbit}/T $
$$ \dfrac{GM_{star} }{(R_{star}+h)^2}= \dfrac{\left(\dfrac{2\pi (R_{star}+h)}{T} \right)^2}{(R_{star}+h)}$$
$$ \dfrac{GM_{star} }{(R_{star}+h)^2}= \dfrac{4\pi^2 (R_{star}+h)^{ \color{red}{\bf\not}2}}{T^2 \color{red}{\bf\not}(R_{star}+h)} $$
Noting that the number of revolutions is given by $N=1/T$
$$N =\dfrac{1}{\sqrt{\dfrac{4\pi^2(R_{star}+h)^3}{GM_{star}}}}$$
Plugging the known;
$$N=\dfrac{1}{\sqrt{\dfrac{4\pi^2([10\times 10^3]+1000)^3}{ (6.67\times 10^{-11})(1.99\times 10^{30})}}}$$
$$N=\bf 1589.35\;\rm rev/s=\color{red}{\bf 9.54\times10^4}\;\rm rev/min$$
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e) According to Kepler's third law;
$$T^2=\left[\dfrac{4\pi ^2}{GM}\right]r^3$$
So the radius of a geosynchronous orbit about the neutron star is given by
$$r^3=\dfrac{T^2 GM_{star}}{4\pi^2}$$
$$r =\sqrt[3]{\dfrac{T^2 GM_{star}}{4\pi^2}}$$
Plugging the known;
$$r =\sqrt[3]{\dfrac{ (1)^2(6.67\times 10^{-11})(1.99\times 10^{30})}{4\pi^2}}$$
$$r=\color{red}{\bf1.498\times 10^6}\;\rm m$$
