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3 years ago

Which statement is true about a planet’s orbital motion?

Physics

1 answer:

lana66690 [7]3 years ago

3 0

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

$V = \sqrt{\frac{GM}{R + h} }$

Where

V - velocity of the orbit at a height h from the surface

R - Radius of the second object

G - Gravitational constant

h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

$1/2 mV^{2} = GMm/R$

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

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The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. What

Ne4ueva [31]

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

$v_1 = v_x$

now when he is at half of the maximum height the speed will be in x and y direction both

$v_2 = \sqrt{v_y^2 + v_x^2}$

here it is given that

$v_1 = 0.58 v_2$

$v_x = 0.58\sqrt{v_x^2 + v_y^2}$

$2.97 v_x^2 = v_x^2 + v_y^2$

$1.97 v_x^2 = v_y^2$

also we know that

$v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}$

here we know that maximum height is given as

$H = \frac{v_{iy}^2}{2g}$

$v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}$

$v_y^2 = \frac{v_{iy}^2}{2}$

now from above

$1.97 v_x^2 = \frac{v_{iy}^2}{2}$

$1.98 v_x = v_{iy}$

also we know that angle of projection is

$tan\theta = \frac{v_{iy}}{v_x}$

$tan\theta = \frac{1.98v_x}{v_x}$

so angle is

$\theta = tan^{-1} 1.98$

$\theta = 63.3 degree$

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