How Would Changes in the Sun's Mass and Earth's Velocity Affect Earth's Orbit?

[ritwik06]

Q.1 If the mass of the sun is reduced by 1% and the Earth's velocity is also reduced by 1%, what will be the trajectory of the earth?

 

[ritwik06]

Q.2 In astronomy, looking deeper into space amounts to looking back into the past. If I want to study an object 1 million years ago compared to a similar object in my neighbourhood, how far in the space should I search?

 

[marcus]

Q.2 In astronomy, looking deeper into space amounts to looking back into the past. If I want to study an object 1 million years ago compared to a similar object in my neighbourhood, how far in the space should I search?

the simple answer is look out approximately 1 million lightyears
but the simple answer glosses over some stuff that you should know.

There are several different ways to quantify distance at astronomical scale. For short distances, like a million LY, the different measures are approximately the same so it is not worthwhile bothering to distinguish. But for longer distances, like a billion LY, the different measures give noticeably different numbers, so it pays to be clear which you are using

Probably the most useful is the present distance----the distance to the object at the present moment in time---which is the distance that occurs in the Hubble Law. It takes account of the fact that space has expanded during the time the light has been traveling.

what I advise you to do is google Ned Wright (a prof at UCLA) and go to his site and find "Cosmology Calculator"
this related different kinds of distance, like redshift, and light travel time, and present (Hubble-Law) distance. you can play around with that and find out what present distance would correspond to a light travel time of a billion years. It will be more than a billion LY because space will have expanded after the light traveled thru it.

 

[marcus]

Q.1 If the mass of the sun is reduced by 1% and the Earth's velocity is also reduced by 1%, what will be the trajectory of the earth?

If you wanted to keep the orbit distance the same, then you would reduce Earth speed by the SQUARE ROOT of the factor you reduce sun mass by.

I know this is not your question but I tell you anyway

the square root of 0.98 is approx 0.99

so if you reduce mass of sun by TWO percent (to 0.98 previous) then in order to keep orbit same size you should reduce speed by ONE percent (to 0.99 of previous speed)
=========================

so in the question you asked, you are reducing the speed by TOO MUCH to keep Earth in present orbit. At the moment you magically reduce sun mass to 0.99 and Earth speed to 0.99, the Earth will enter a new orbit which is more elliptical and makes it fall in closer to the sun, a little bit.

After it has gone around the sun it will come back to about the same point where you made the magic change, but its orbit will have been more of an oval and during part it would have been closer than before.

you should not have done that, people will not like getting hotter for part of the year

if you wanted to reduce sun mass by one percent, then you should have reduced Earth speed by HALF a percent
because the square root of 0.99 is approximately 0.995

 

[Labguy]

Q.1 If the mass of the sun is reduced by 1% and the Earth's velocity is also reduced by 1%, what will be the trajectory of the earth?

The Sun would still be considered a "massive body", especially relative to the Earth's mass. Therefore, reducing the orbital velocity of the Earth would make that velocity less than necessary to maintain orbit at present or lesser distance so the Earth would slowly spiral inward and eventually decay to where it would spiral into the Sun.

 

[tony873004]

I agree with Marcus.

If you wanted to spell it out, formula for circular velocity is: V=sqr(GM/r)

So for the Earth in its present configuration is
sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)

And under a Sun 0.99 times as massive as the current Sun, to maintain a circular orbit at the same distance would be
sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)

To compute the ratios:
sqr(6.67e-11*(1.989e30*.99+5.97e24)/149597870691) / sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)=0.994987

Notice that Everything but the 0.99 under the root symbol in the numerator cancels out, leaving you with sqr(0.99)=0.994987

 

[Labguy]

I agree with Marcus.

If you wanted to spell it out, formula for circular velocity is: V=sqr(GM/r)

So for the Earth in its present configuration is
sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)

And under a Sun 0.99 times as massive as the current Sun, to maintain a circular orbit at the same distance would be
sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)

To compute the ratios:
sqr(6.67e-11*(1.989e30*.99+5.97e24)/149597870691) / sqr(6.67e-11*(1.989e30+5.97e24)/149597870691)=0.994987

Notice that Everything but the 0.99 under the root symbol in the numerator cancels out, leaving you with sqr(0.99)=0.994987

Ok, but Marcus said it would change to a more elliptical orbit and you give the formulae for a circular, closer orbit. Wouldn't it do a more eccentric ellipse like Marcus posted? And, would it ever return to the aphelion radius we now have??

Every time I post anything re: solar system/planetary/orbital stuff I always screw it up..
I just got to stick with stars, supernovae, black holes, etc. At least sometimes I get some of those right...

 

[SpaceTiger]

Ok, but Marcus said it would change to a more elliptical orbit and you give the formulae for a circular, closer orbit. Wouldn't it do a more eccentric ellipse like Marcus posted? And, would it ever return to the aphelion radius we now have??

If we approximate the current orbit as circular, then yes, it ought to enter an elliptical orbit with aphelion at the current earth-sun distance.