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2 years ago

A projectile is fired vertically from Earth's surface with

Physics

1 answer:

scoray [572]2 years ago

5 0

Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let m = Mass of the projectile

Solution:

Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

$-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2$

$-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )$

= $( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$

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Read 2 more answers

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$