Solving for Mercury's Perihelion Distance

  • Thread starter Thread starter Kinash
  • Start date Start date
AI Thread Summary
To solve for Mercury's perihelion distance, the discussion emphasizes the relationship between potential energy (Pe) and kinetic energy (Ke), noting that their sum remains constant. The problem states Mercury's orbital speeds at aphelion and perihelion, along with its distance from the sun at aphelion. The relevant equations include the gravitational potential energy formula and the concept of areal velocity from Kepler's Law, which maintains a constant value during the orbit. The participants seek clarification on the formula for areal velocity to aid in their calculations. Understanding these principles is crucial for determining Mercury's distance from the sun at perihelion.
Kinash
Messages
1
Reaction score
0
I have been struggling with this particular problem related to gravity.

Homework Statement


Mercury's orbital speed varies from 38.8 km/s at aphelion to 59.0 km/s at perihelion. If the planet is 6.99x10^10m from the sun's center at aphelion, how far is it at perihelion.


Homework Equations



Pe+Ke=constant

Pe= (-Gm1m2)/r2


Sun mass = 1.99x1030kg
Sun Mean Radius= 696x10m
Sun surface gravity=274m/s2
Sun Mean distance from central body=2.6x1017km

Mercury mass = .330x1024kg
Mercury mean Radius = 2.44x106m
Mercury Surface Gravity = 3.7m/s2
Mercury Mean dist. from central body = 57.6x106km
 
Physics news on Phys.org
According to Kepler's Law, when a planet is orbiting around the sun its areal velocity remains constant. So what is the formula for the areal velocity?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Variation of gravitational attraction between Sun and Earth
Replies
3
Views
2K
Calculating Orbital Speed Using Conservation of Angular Momentum
Replies
2
Views
1K
Period of a planet orbiting the Sun
Replies
11
Views
2K
Calculate Mercury's Orbital Speed and Distance: Universal Gravity Equation
Replies
6
Views
4K
Time to Mars (orbit around Sun given perihelion and aphelion)
Replies
2
Views
3K
Orbital Velocity Calculation for an Asteroid at Perihelion
Replies
1
Views
2K
Calculate The Surface Temperature of Mercury
Replies
5
Views
10K
How Can Kepler's 3rd Law and Related Equations Be Applied to Orbital Mechanics?
Replies
6
Views
3K
I Atmospheric escape parametre statistics
Replies
1
Views
2K
Calculating Gravity and Orbital Speed of Mercury at Mean Distance from the Sun
Replies
10
Views
4K

Hot Threads

Recent Insights

Back
Top