All categories

3 years ago

What is the orbital period in years of a planet with a semi major axis of 35 au

Physics

1 answer:

mezya [45]3 years ago

6 0

Answer:

Orbital period of the planet will be 207.06 year

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third law there is relation between the orbital period and semi major axis which is t $T^2=R^3$

So $T^2=35^3$

$T^2=42875$

$T=207.06year$

So orbital period of the planet will be 207.06 year

You might be interested in

Assuming motion is on a straight path, what would result in two positive components of a vector?

erastovalidia [21]

Assuming motion is on a straight path, the result of two positive components of a vector would also be a positive value since both are having positive signs and directions. The direction would be the same with the motion as well. Hope this answers the question. Have a nice day.

3 0

3 years ago

Read 2 more answers

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window w

expeople1 [14]

The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity u from a window, which is at a height y from the ground. The ball lands at a distance x from the base of the building. Let the ball take a time t to reach the ground. In the time t ,the ball falls a vertical distance y and also travel a horizontal distance x.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

$y=\frac{1}{2} gt^2$