# If the height of a satellite completing one revolution around the earth in T seconds is h1 meter,

If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking $$2\sqrt{2}T$$ seconds for one revolution?

## 1 Answer

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Time period of the satellite is given as
$$T = \frac{2π(R+h)^{3/2}}{\sqrt{GM}}$$

When the height of the satellite is h1, it takes T time to revolve around the Earth. Thus, at height h1
$$T = \frac{2π(R+h_1)^{3/2}}{\sqrt{GM}}$$ .... (i)

When the satellite takes $$2\sqrt{2}T$$ time to revolve around the Earth, let it be at height h2. Thus,
$$2\sqrt{2}T =\frac{2π(R+h_2)^{3/2}}{\sqrt{GM}}$$ .....(ii)

Dividing (ii) by (i), we get
$$2\sqrt{2} =\frac{(R+h_2)^{3/2}}{(R+h_1)^{3/2}}$$

$$2^{3/2} =\frac{(R+h_2)^{3/2}}{(R+h_1)^{3/2}}$$

$$2 =\frac{(R+h_2)}{(R+h_1)}$$

2R+2h1 = R+h2

∴ h2=R+2h1

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