Physics: In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 7.08 × 103 s and orbital speed of 3.40 × 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106 m. Calculate the mass of Mars.

A shows a balance force while B shows an unbalance...

In March 1999 the Mars Global Surveyor (GS) entered

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17.03.2022

$5.944\times 10^{23}\ kg$

Explanation:

Given

Time period $T=7.08\times 10^3\ s$

Orbital speed $v=3.40\times 10^3\ m/s$

mass of GS $m_{GS}=930\ kg$

Radius of Mars $r=3.43\times 10^6\ m$

Consider the mass of mars is M

Here, Gravitational pull will provide the centripetal force

F_G=F_c

$\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}$

$M=5.944\times 10^{23}\ kg$

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