Attractive force between Venus and Sun

[sadifermi]

I just need someone to double check my answer and confirm they get the same answer as I do. Thanks.
Question: Find the attractive force between Vneus and the sun. The distance between their centers is 1.08x10^11.

Equation: F = Gm1m2/d^2
G= 6.67x10^-11 N.m^2/kg^2
m1 (mass of sun)= (1.98x10^30kg)
m2 (mass of venus) = (4.83x10^24kg)
d= (1.08x10^11)

= (6.67x10^-11 N.m^2/kg^2)(1.98x10^30kg)(4.83x10^24kg)/(1.08x10^11)^2
=5.4687 x 10^22

Please let me know if this is the correct answer asap. Thanks!

 

[Borek]

Your answer has no units.

 

[PeterO]

I just need someone to double check my answer and confirm they get the same answer as I do. Thanks.
Question: Find the attractive force between Vneus and the sun. The distance between their centers is 1.08x10^11.

Equation: F = Gm1m2/d^2
G= 6.67x10^-11 N.m^2/kg^2
m1 (mass of sun)= (1.98x10^30kg)
m2 (mass of venus) = (4.83x10^24kg)
d= (1.08x10^11)

= (6.67x10^-11 N.m^2/kg^2)(1.98x10^30kg)(4.83x10^24kg)/(1.08x10^11)^2
=5.4687 x 10^22

Please let me know if this is the correct answer asap. Thanks!

You have just omitted the next digits rather than round off to the 5 figures you have chosen to specify.

besides, your answer should only be to 3 figures!

And as per the previous post - you need some units.

 

[sadifermi]

Thanks folks, appreciate it!

 

[mfb]

WolframAlpha gets a slightly different result: 5.54*10^22 N

It uses a mass of venus which is larger by 1% and a mass of sun which is larger by .5%.

 

[PeterO]

Thanks folks, appreciate it!

So you should have been saying 5.47 (correctly rounded) not 5.46 (merely truncated)