Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago

If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?

julsineya [31]

Yes, the above-given statement is true

<u>Explanation:</u>

The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
Momentum is described as the mass of the object multiplied by its velocity.
<u>Momentum (p) = Mass (M) * Velocity (v)</u>
Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

4 0

3 years ago