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Shane Macguire Grade: upto college level
`        A particle is fired vertically upward from earth’s surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle. `
3 years ago

396 Points
```										Sol. The particle attain maximum height = 6400 km.
On earth’s surface, its P.E. & K.E.
E base e (1/2) mv^2 + (-GMm/R) …(1)
In space, its P.E. & K.E.
E base s (GMm/R + h) + 0
E base s (-GMm/R + h)  …(2) (∴ h = R)
Eqating (1) & (2)
-  GMm/R + 1/2 mv^2 = GMm/2R
Or (1/2) mv^2 = GMm (-1/2R + 1/R)
Or v^2 = GM/R
= 6.67 * 10^-11 * 6 * 10^24/6400 * 10^3
= 40.02 * 10^13/6.4 * 10^6
= 6.2 * 10^7 = 0.62 * 10^8
Or v = √0.62 * 10^8 = 0.79 * 10^4 m/s = 7.9 km/s.

```
3 years ago
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