Earth WUUIU Uoriginal mass?Ans: ~ 12 hrsIf the height of a satellite c

1.	Earth WUUIU Uoriginal mass?Ans: ~ 12 hrsIf the height of a satellite completinone revolution around the earth inseconds is h, meter, then what woulbe the height of a satellite taking2/2 T seconds for one revolution?Ans: R+ 2hject :
Answer» Answer: plz mark as brainliest and follow me please Explanation:•Kepler's Third LAW: The cube of the MEAN distance of a planet from the sun is directly proportional to the square of time TAKEN to move around the sun. where , r is the mean distance from the sun •°•°•°•°•°<><><<><>><><>°•°•°•°•°•° ¶¶¶ SOLUTION: Given, height of satellite = m time taken by satellite for one revolution around earth = T sec When satellite takes seconds for one revolution, height of satellite = ? Cross Multiply 8(R+h_{1})^{3} = (R+h_{2})^{3} \implies 2^{3}(R+h_{1})^{3} = (R+h_{2})^{3} \implies (2(R+h_{1}))^{3} = (R+h_{2})^{3} \implies (2R+2h_{1})^{3} = (R+h_{2})^{3} APPLYING cuberoot on both sides, we get \implies 2R+2h_{1} = R+h_{2} \implies h_{2} = 2R+2h_{1}-R \implies h_{2} = R+ 2h_{1} •°• If satellite takes 2\sqrt{2}T seconds for one revolution, then height of satellite = R+2h_{1}

Earth WUUIU Uoriginal mass?Ans: ~ 12 hrsIf the height of a satellite completinone revolution around the earth inseconds is h, meter, then what woulbe the height of a satellite taking2/2 T seconds for one revolution?Ans: R+ 2hject :

Answer»

Answer: plz mark as brainliest and follow me please

Explanation:•Kepler's Third LAW:

The cube of the MEAN distance of a planet from the sun is directly proportional to the square of time TAKEN to move around the sun.

where , r is the mean distance from the sun

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°

¶¶¶ SOLUTION:

Given,

height of satellite = m

time taken by satellite for one revolution around earth = T sec

When satellite takes seconds for one revolution, height of satellite = ?

Cross Multiply

8(R+h_{1})^{3} = (R+h_{2})^{3}

\implies 2^{3}(R+h_{1})^{3} = (R+h_{2})^{3}

\implies (2(R+h_{1}))^{3} = (R+h_{2})^{3}

\implies (2R+2h_{1})^{3} = (R+h_{2})^{3}

APPLYING cuberoot on both sides, we get

\implies 2R+2h_{1} = R+h_{2}

\implies h_{2} = 2R+2h_{1}-R

\implies h_{2} = R+ 2h_{1}

•°• If satellite takes 2\sqrt{2}T seconds for one revolution, then height of satellite = R+2h_{1}