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bezimeni [28]
2 years ago
9

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

A orbits at an altitude of 4500 km , and satellite B orbits at an altitude of 11100 km .
c) How much work would it require to change the orbit of satellite A to match that of satellite B?
Physics
1 answer:
-Dominant- [34]2 years ago
7 0

Answer:

The required work done is 6.5\times10^{9}\ J

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}

Put the value into the formula

U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}

U_{A}=-3.44\times10^{10}\ J

We need to calculate the potential energy

Using formula of potential

U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}

Put the value into the formula

U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}

U_{B}=-2.14\times10^{10}\ J

We need to calculate the value of k_{A}

Using formula of k_{A}

k_{A}=-\dfrac{1}{2}U_{A}

Put the value into the formula

k_{A}=\dfrac{1}{2}\times3.44\times10^{10}

k_{A}=1.72\times10^{10}\ J

We need to calculate the value of k_{B}

Using formula of k_{B}

k_{B}=-\dfrac{1}{2}U_{B}

Put the value into the formula

k_{B}=\dfrac{1}{2}\times2.14\times10^{10}

k_{B}=1.07\times10^{10}\ J

We need to calculate the work done

Using formula of work done

W=\Delta K+\Delta U

W=(k_{B}-k_{A})+(U_{B}-U_{A})

W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})

W=\dfrac{1}{2}(U_{B}-U_{A})

Put the value into the formula

W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})

W=6.5\times10^{9}\ J

Hence, The required work done is 6.5\times10^{9}\ J

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